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k • 



MATHEMATICAL TEXTS 

Edited by 

Percey F. Smith, Ph.D. 

Professor of Mathematics in the Sheffield Scientific School of 

Yale University 

Elements of Differential and Integral Calculus 
By W. A. Granville, Ph.D. 

Elements of Analytic Geometry 

By P. F. Smith and A. S. Gale, Ph.D. 

Introduction to Analytic Geometry 

By P. F. Smith and A. S. Gale, Ph.D. 

Theoretical Meclianics In press 

By P. F. Smith and O. C. Lester, Ph.D. 

Advanced Algebra 

By H. £. Hawkes, Ph.D. 



Strength of Materials 

By S. E. Slocum, Ph.D., and E. L. Hancock, M.Sc. 



ELEMENTS OF THE DIFFERENTIAL 
AND INTEGRAL CALCULUS 



B7 



WILLIAM ANTHONY GRANVILLE, Ph.D. 

nmTRUOTOiR or MATHBMATirS IK THE SHEFFIELD SCISNTIFIO SCHOOL 

Tale Univebsity 



With the Editorial Cooperation of 



PERCEY F. SMITH, Ph.D. 

pbofe880b of mathematics ik the sheffield soibktifio school 

Yale Ukivebsity 




f Dm ( A] |f_\" 

(if -^i, -^, -%>-. .t, ^^ ^^ .;j;^, ,.^^^ .. -, , ^ . , -.^ .J 









GINN & COMPANY 

BOSTON • NEW YORK • CHICAGO • LONDON 



\ 

. V 
I 



623251 



COPTRI^p, 1904, BY 
WnAAAM AlTTHONY QBAMVILLE AND PESCEY F. SMITB 



ALL RIGHTS BB8KBYXD 
27.3 



CINN & COMPANY • PRO- 
PRIETORS • BOSTON • U^Jk. 



PEEFACE 

The present volume is the result of an effort to write a modem 
text-book on the Calculus which shall be essentially a drill book. 
With this end in view, the pedagogic principle that each resiQt 
should be made intuitionally as well as analytically evident to the 
student has been kept constantly in mind. Indeed it has been 
thought best in some cases, as for example in Maxima and Minima 
and the Theorem of Mean Value, to discuss the question first 
from the intuitional side, in order that the significance of the new 
idea might be made plain in the ^ost direct manner. The object 
has not been to teach the student to rely upon his intuition^ but in 
some cases to use this facility in advance of the analytical investi- 
gation. The short chapter on Numbers is intended to give the 
student a chance to review his ideas of number. Limits and 
Continuity are treated at length, the latter mostly from a graphical 
standpoint, — the only method suited to a first course. In fact, 
graphical illustration has been di^wn upon to the fullest extent 
throughout the book. 

As special features, attention may be called to the effort to 
make perfectly clear the nature and extent of each new theorem, 
the large number of carefully graded exercises, and the summa- 
rizing into working rules of the methods of solving problems. In 
the Integral Calculus the notion of integration over a plane area 
has been much enlarged upon, and integration as the limit of a 
summation is constantly emphasized. The book contains more 
material than is necessary for the usual course of one hundred 
lessons given in our colleges and engineering schools; but this 
gives teachers an opportunity to choose such topics as best suit 
the needs of their classes. It is believed that the volume contains 
all subjects from which a selection naturally would be made in 
preparing students either for elementary work in applied sciencQ 
or for more advanced work in pure mathematiQs, 

iU 



iv PREFACE 

Certain proofs of considerable difficulty (as the existence of the 
number e) have been inserted with the belief that, while it is not 
always advisable to require beginners to learn them, a discussion 
of them with the class will render such investigations profitable 
and stimulating. 

With a few exceptions the author has found it impracticable to 
acknowledge his indebtedness to the large number of American, 
English, and continental writers whose books and articles have 
helped and inspired him in the work, the bulk of the matter 
having long been the common property of all mankind. While 
many of the exercises are new, a large number are standard and 
are to be found in many of the best treatises. 

The author's acknowledgments are due to Professor M. B. 
Porter of the University of Texas for critically examining the 
manuscript, to Professor James Pierpont of Yale University for 
many valuable suggestions, to my former colleagues. Professor 
E. R. Hedrick of the University of Missouri and Dr. C. N. 
Haskins, for their interest and assistance, to Dr. C. E. Stromquist 
of the University of Princeton for verifying the examples, and to 
my colleague, Mr. L. C. Weeks, for drawing the figures. The 
thanks of the author are also due to his former instructor in mathe- 
matics. Professor John E. Clark, now Professor Emeritus of Yale 
University, who first advised and encouraged him to undertake 
the task of writing this book. 

Sheffield Scientific School, 
Tale Univer8itt, July, 1904. 



CONTENTS 



DIFFERENTIAL CALCULUS 



Chapter I 
COLLECTION OF FORMULAS 

SEOnOK PAOX 

1. Formulas for reference from Algebra, Trigonometry, and Analytic 

Geometry « 1 

2. Greek alphabet 4 

3. Natural values of trigonometric functions . * . . • 4 

4. Rules for signs in the four quadrants 6 



Chapter II 

NUMBERS 

5. Rational numbers 7 

6. Comparison of rational numbers with the points of a straight line . 7 

7. Irrational numbers 8 

8. Real numbers 8 

9. Numerical or absolute value 

10. Imaginary numbers 9 

11. Complex numbers 9 

12. Division by zero excluded 10 

13. Only real numbers considered 10 

Chapter III 

VARIABLES AND FUNCTIONS 

14. Variables . 11 

15. Constants 11 

16. Interval of a variable 11 

17. Continuous variation 11 

18. Functions 12 

19. Independent and dependent variables . . . . . .12 

20. Notation of functions 13 

21. Values of the independent variable for which a function is defined . 14 

V 



vi CONTENTS 

SECTION PAGE 

22. One-valued and many-yalued functions ...... 14 

23. Explicit functions 15 

24. Inyerse functions 15 

25. Integral rational functions 16 

26. Rational functions 16 

27. Explicit algebraic functions 17 

28. Transcendental functions 17 

Chapter IV 

THEORY OF LIMITS 

20. Limit of a variable 19 

80. Infinitesimals 21 

31. The concept of infinity (oo) 21 

32. Limiting value of a function 22 

33. Continuous and discontinuous functions 22 

34. Continuity and discontinuity of functions illustrated by their graphs 23 

35. Fundamental theorems on limits 26 

36. Special limiting values .29 

37. The number e .31 

Chapter V 

DIFFERENTIATIOlf 

38. Introduction 37 

39. Increments 37 

40. Comparison of increments 38 

41. Derivative of a function of one variable 39 

42. Symbols for derivatives 40 

43. Differentiable functions 41 

44. General rule for differentiation ....... 42 

45. Applications of the derivative to Greometry 43 

Chapter VI 
RULES FOR DIFFERENTIATING STANDARD ELEMENTARY FORMS 

46. Importance of general rule 46 

47. Differentiation of a constant 48 

48. Differentiation of a variable Yiiih respect to itself .... 49 

49. Differentiation of a sum 49 

50. Differentiation of the product of a constant and a variable . . 50 



CONTENTS vii 

BEOnOir PAGE 

51. Differentiation of the product of two variables .... 50 

52. Differentiation of the product of any finite number of variables . 50 

53. Differentiation of a variable with a constant exponent ... 51 

54. Differentiation of a quotient 52 

55. Differentiation of a function of a function 57 

56. Differentiation of inverse functions 58 

57. Differentiation of a logarithm 58 

58. Differentiation of the simple exponential function .... 60 

59. Differentiation of the general exponential function ... 61 

60. Logarithmic differentiation . . ' 62 

61. Differentiation of sin v 67 

62. Differentiation of coav 68 

63. Differentiation of tan v 68 

64. Differentiation of cot o 69 

65. Differentiation of sec v 69 

66. Differentiation of CSC . 70 

67. Differentiation of vers v 70 

68. Differentiation of arc sin v 74 

69. Differentiation of arc cos <; 75 

70. Differentiation of arc tan v 76 

71. Differentiation of arc cot© 77 

72. Differentiation of arc sec t; 77 

73. Differentiation of arc esc t; 78 

74. Differentiation of arc vers v 78 

75. Implicit functions 83 

76. Differentiation of implicit functions 84 

Chapter VII 
SIMPLE APPLICATIONS OF THE DERIVATIVB 

77. Direction of a curve 86 

78. Equations of tangent and normal, lengths of subtangent and sub- 

normal. Rectangular coordinates 89 

79. Parametric equations of a curve 92 

80. Angle between the radius vector drawn to a point on a curve and 

the tangent to the curve at that point 97 

81. Lengths of polar subtangent and polar subnormal 98 

82. Solutions of equations having multiple roots 100 

83. Applications of the derivative in mechanics. Velocity . . .102 

84. Component velocities ^^^ 

85. Acceleration 1^5 

86. Component accelerations . . . • • • • .105 



viu CONTENTS 



Chapter Vm 
SUCCESSIVE DIFFEREHTUTIOH 

SECTION PAOB 

87. Definition of Buccessive derivatives 109 

88. Notation 100 

89. The nth derivative 110 

90. Leibnitz's formula for the nth derivative of a product . . .111 

91. Successive differentiation of implicit functions . . . .112 



Chapter IX 
MAXIMA AHD MINIMA 

92. Increasing and decreasing functions 116 

93. Tests for determining when a function is increasing and when 

decreasing 117 

94. Maximum and minimum values of a function .... 118 

95. First method for examining a function for maximum and mini- 

mum values 121 

96. Second method for examining a function for maximum and mini- 

mum values 123 

97. General directions for solving problems involving maxima and 

minima 120 



Chapter X 
POINTS OF INFLBCTION 

98. Definition of points of inflection and rule for finding points of 

inflection 136 

Chapter XI 
DIFFERENTULS 

99. Introduction . 140 

100. Definitions ....;..... 140 

101. dx and dy considered as infinitesimals 141 

102. Derivative of the arc in rectangular codrdinates .... 141 

103. Derivative of the arc in polar codrdinates 143 

104. Formulas for finding the differentials of functions . . . 144 

105. Successive differentials ......... 146 



CONTENTS ix 



Chapter XII 
RATES 

SECTION PAQB 

106. The derivative considered as the ratio of two rates • • .148 



Chapter XIII 
CHANGE OF VARIABLE 

107. Interchange of dependent and independent variables . . .152 

108. Change of the dependent variable 153 

109. Change of the independent variable 154 

110. Simultaneous change of both independent and dependent variables 156 



Chapter XIV 
CURVATURE. RADIUS OF CURVATURE 

111. Curvature '. . . 159 

112. Curvature of a circle 159 

113. Curvature at a point 160 

114. Formulas for curvature 161 

115. Radius of curvature 162 



Chapter XV 
THEOREM OF KEAN VALUE. DTDETERMinATE FORMS 

116. Rolle's Theorem 166 

117. The Theorem of Mean Value 167 

118. The Extended Theorem of Mean Value 168 

119. Maxima and minima treated analytically 169 

120. The Generalized Theorem of Mean Value 172 

121. Indeterminate forms 172 

122. Evaluation of a function taking on an indeterminate form . 173 

123. Evaluation of the indeterminate form - 174 

124. Evaluation of the indeterminate form — 177 

00 

125. Evaluation of the indeterminate form • oo 179 

126. Evaluation of the indeterminate form oo — oo . . . .179 

127. Evaluation of tJie indeterminate forms 0*, 1*, oo^ . . . 181 



CONTENTS 



Chapteb XYI 
circlb of curvature. center of curvature 

SEOnOK PAOB 

128. Circle of curvature. Center of curvature 183 

129. Center of curvature the limiting position of the intersection of 

normals at neighboring points 186 

130. Evolutes 187 

131. Properties of the evolute 190 

132. Involutes and their mechanical construction 191 



Chapter XVn 
PARTIAL DIFFEREHTUTIOH 

133. Continuous function of two or more independent variables . .193 

134. Partial derivatives 194 

135. Partial derivatives interpreted geometrically 195 

. 136. Total derivatives 198 

137. Total differentials 200 

138. Differentiation of implicit functions . 1 . • . . 202 

139. Successive partial derivatives 203 

140. Order of differentiation immaterial 206 

Chapter XVIII 
ENVELOPES 

141. Family of curves. Variable parameter 208 

142. Envelope of a family of curves depending on one parameter . . 208 

143. The envelope touches each curve of the family at the limiting 

points on the curve 210 

144. Parametric equations of the envelope of a family depending on 

one parameter . . . 211 

145. The evolute of a given curve considered as the envelope of its 

normals 213 

146. Two parameters connected by one equation of condition • .214 

Chapter XIX 
SERIES 

147. Introduction 217 

148. Infinite series 218 

149. Existence of a limit 220 

150. Fundamental test for convergence ...'••. 221 



CONTENTS xi 

SXCnON PAOB 

151. Comparison test for convergence 222 

152. Cauchy's ratio test for convergence 224 

153. Alternating series . . ' 226 

154. Absolute convergence 227 

155. Power series 228 

Chapter XX 
EXPANSION OF FUNCTIONS 

156. Introduction 231 

157. Taylor's Theorem and Taylor's Series 232 

158. Maclaurin's Theorem and Maclaurin's Series .... 234 

159. Computation by series 238 

160. Approximate formulas derived from series 240 

161. Taylor's Theorem for functions of two or more variables . . 243 

162. Maxima and minima of functions of two independent variables . 246 

Chapter XXI 
ASYMPTOTES. SINGULAR POINTS. CURVE TRACING 

163. Rectilinear asymptotes 252 

164. Asymptotes found by method of limiting intercepts . . . 252 

165. Method for determining asymptotes to algebraic curves . . 253 

166. Asymptotes in polar coordinates 257 

167. Singular points 259 

168. Determination of the tangent to an algebraic curve at a given 

point by inspection 259 

169. Nodes 262 

170. Cusps 263 

171. Conjugate or isolated points 264 

172. Transcendental singularities 265 

173. Curve tracing 266 

174. General directions for tracing a curve whose equation is given in 

rectangular coordinates 267 

175. Tracing of curves given by equations in polar coordinates . . 269 

Chapter XXII 
APPLICATIONS TO GEOMETRY OF SPACE 

176. Tangent line and normal plane to a skew curve whose equations 

are given in parametric form 271 

177. Tangent plane to a surface 273 



xii CONTENTS 

asonoir paob 

178. Normal line to a Emifaoe 275 

179. Another form of the equations of the tangent line to a skew curve 277 

180. Another form of the equation of the normal plane to a skew curve 277 

Chapter XXm 
CURVES FOR REFERENCE 



INTEGRAL CALCULUS 

Chapter XXIV 
INTEGRATION. RULES FOR INTEGRATING STANDARD ELEMENTARY FORMS 

181. Integration 287 

182. Constant of integration. Indefinite integral . . . .289 

183. Rules for integrating standard elementary forms .... 291 

184. Trigonometric differentials 303 

Chapter XXV 
CONSTANT OF INTEGRATION 

185. Determination of the constant of integration by means of initial 

conditions 309 

186. Geometrical signification of the constant of integration . . 309 

187. Physical signification of the constant of integration . . • 310 

Chapter XXVI 
INTEGRATION OF RATIONAL FRACTIONS 



188. Introduction . 315 

189. Partial fractions 315 

190.. Imaginary roots 318 

191. Case I 320 

192. Caaell 321 

193. Casein 322 

194. Case IV 324 



CONTENTS xiii 



Chaptbr xxvn 

niTEGRATIOH BT SUBSTITUTION OP A HEW VARIABLE. 
,,_^„ RATIONALIZATIOE 



195. Introduction , , 

196. Differentials containing fractional powers of x only 

197. Differentials containing fractional powers oi a-\- hx only 

198. Differentials containing no radical except Va + ftz + x* 

199. Differentials containing no radical except Vo + fti — ^ 

200. Binomial differentials 

201. Conditions of integrability of binomial differentials 

202. Transformation of trigonometric differentials 

203. MisceUaneous substitutions 



329 
329 
330 

331 

332 
334 
334 
337 
339 



, Chapter XXVIII 
INTEGRATION BT PARTS. REDUCTION FORMULAS 

204. Formula for integration by parts 341 

205. Reduction formulas for binomial differentials .... 344 

206. Reduction formulas for trigonometric differentials . . . 349 

207. To find fc^sin nxdx and (e^ cos nxdx 353 

Chapter XXIX 
THE DEPINITE INTEGRAL 

208. Differential of an area . . . ., 355 

209. The definite integral 356 

210. Geometrical representation of an integral 357 

211. Mean value of 0(z) 358 

212. Interchange of limits 358 

213. Decomposition of the interval 359 

214. The definite integral a function of its limits .... 359 

215. Calculation of a definite integral 360 

216. Infinite limits 360 

217. When (x) is discontinuous 361 

218. Change in limits 365 

Chapter XXX 
INTEGRATION A PROCESS OF SUMMATION 

219. Introduction 367 

220. The definite integral as the limit of a sum 367 



xiv CONTENTS 

SBOTIOK PAGE 

221. Areas of plane curves. Rectangular codrdinates • • • .871 

222. Area when curve is given in parametric form .... 373 

223. Areas of plane curves. Polar codrdinates 876 

224. Length of a curve 378 

225. Lengths of plane curves. Rectangular codrdinates . . . 379 

226. Lengths of plane curves. Polar codrdinates 382 

227. Volumes of solids of revolution 384 

228. Areas of surfaces of revolution 388 

Chapter XXXI 
SUCCESSIVE AND PARTIAL Of TEGRATIOV 

229. Successive integration 892 

230. Partial integration 394 

231. Definite double integral. Greometric interpretation . . . 396 

232. Value of a definite double integral over a region .... 400 

233. Plane area as a definite double integral. Rectangular codrdinates 402 

234. Plane area as a definite double integral. Polar codrdinates . . '406 

235. Moment of inertia. Rectangular codrdinates .... 407 

236. Moment of inertia. Polar codrdinates 410 

237. General method for finding the areas of surfaces .... 411 

238. Volumes found by triple integration 415 

239. Miscellaneous applications of the Integral Calculus . . .419 

Chapter XXXII 
ORDDTART DIFFERENTIAL EQUATIOHS 

240. Differential equations. Order and degree 424 

241. Solutions of differential equations 425 

242. Verifications of solutions 426 

243. Differential equations of the first order and of the first degree . 427 

244. Differential equations of the nth order and of the first degree . 435 

Chapter XXXIH 
DTTEGRAPH. TABLE OF INTEGRALS 

245. Mechanical integration 446 

246. Integral curves . ' • . . 446 

247. The integraph ' . . .448 

248. Integrals for reference 450 

INDEX 461 



DIFFERENTIAL CALCULUS 



CHAPTER I 

COLLECTION OF FORMULAS 

1. Formulas for reference. For the convenience of the student 
we give the following list of elementary formulas from Algebra, 
Geometry, Trigonometry, and Analytic Geometry. 

1. Binomial Theorem (n being a positive integer) : 

I? l? 

lr-1 



Also written : 

2. nI=[n = l-2-8-4---(n-l)n. 

3. In the quadratic equation ox^ + &x + c = 0, 

when &3 — 4 oc > 0, the roots are real and unequal ; 
when &^ — 4 oc = 0, the roots are real and equal ; 
when &3 — 4 oc < 0, the roots are imaginary. 

4. When a quadratic equation is reduced to the form x> + jpz = 9, 

p = sum of roots with sign changed, 
and q = product of roots with sign changed. 

6. In an arithmetical series, 

i = a + (n-l)d;« = ?(a + = |[2a + (n-l)(q. 

6. In a geometrical series, 

, rl-a a(r*-l) 

r — 1 r — 1 

1 



DIFFERENTIAL CALCULUS 

7. log a& = log a + log 5. 9. log ofl^n log a. 

8. log - = log a ~ log 6. 10. log Va = - log a, 

11. log 1=0. 12. logaa = l. 13. log- = - logo. 

14. Circamference of circle = 2 srr.* 16. Volame of prism = Ba, 

16. Area of circle = fcr^. 17. Volume of pyramid = i Ba. 

18. Volume of right circular cylinder = icr^. 

19. Lateral surface of right circular cylinder = 2 icra. 

20. Total surface of right circular cylinder = 2 nrr (r + a). 

21. Volume of right circular cone = ^ icr^a, 

22. Lateral surface of right circular cone = ma. 

23. Total surface of right circular cone = jrr (r + s), 

24. Volume of sphere = | m*. 26. Surface of sphere = 4 itr^. 

26. sin X = ; cos x = ; tan z = 



cscx secx cotx 

«_- ^ sinx. ^ cosx 

27. tanx= ;cotx = -; 

cos X sm X 

28. sin«x + cos'x = 1 ; 1-f tan^x = sec«x ; 1 + cot«x = C8c«x. 

29. sin X = COB ( x j ; 30. sin (n — x) = sin x ; 

C08X = sln ^— -x^; cos (jr — x) = — cosx; 

tanx = cot (--«)• ^°(* - X) = - tanx. 

31. sin (x + y) = Bin X cosy + cosx sin y. 

32. sin (x ~ y) = sin X cos y — cosx sin y. 

33. cos (x + y) = cosx cosy — sin X sin y. 

tan X 4- tan V ^. tanx — tan v 

34. tan(x-f y)= ^*^*^ ^ . 36. tan(x-y)= ^ 



1 — tan X tan y 1 + tan x tan y 

2 tanx 



36. sin 2x = 2 sin X cos X ; cos 2x = cos^x — sin^x ; tan 2x = 



l-tan^x 



X 

2tan- 

XX XX 2 

37. sin X = 2 sin - cos - ; cos x = cos^ — sin^ - ; tan x = 



2 2' 2 2 ,^^,x 

2 
<* In formulas 14-26, r denotes radius, a altitude, B area of base, and s slant height. 



COLLECTION OF FORMULAS 3 



38. co«»« = - + -coB2a5: 8m*a5 = cob2x. 

2 2 2 2 

39. l + co8X = 2co8>-: l-ooBX = 28m3-. 

2 2 

^rt . « /I ~ COS X X /l-fcosx ^ « /r^ 



— C08X 



COBS 

41. sin X + sin y = 2 sin i (2 + y) COS i (x — y). 

42. sin X — sin y = 2 cos i (x + y)Bmi{x — y), 

43. cosx + cosy = 2cosi(x + y)cosi(x — 2^). 

44. cos X — cos y = — 2 siu i (x + y) sin i (x — y), 

46. = = : Law of Sines. 

sin ^ sin J3 sin C 

46. (j^ = W + c2 - 2 6c cos -4 ; Law of Cosines. 

47. d = V(xi — x«)2 + (yi — y2)' ; distance between points (Xi, yi) and (x», yj). 

48. d = — ?L±_^L±_.; distance from line -4x + By + C = to (xi, yi). 

49. X = ?l^t_?? , y = Vl±yi . coordinates of middle point. 

60. X = xo + x', y = yo + y' ; transforming to new origin (xo, yo)* 

61. X = x' cos — y' sin 0, y = x' sin + / cos $ ; transforming to new axes mak- 
ing the angle $ with old. 

62. X = p cos $, y = pa\n0; transforming from rectangular to polar coordinates. 

63. p = VxM-T^i ^ = arc tan - ; transforming from polar to rectangular coor- 
dinates. ^ 

64. Different forms of equation of a straight line : 
(a) ILzyi = yt-Vi ^ two-point form ; 

X — Xj Xj — Xi 
X t/ 

(h) - -f I = 1, intercept form ; 
a 

(c) y — yi = m (x — Xi), slope-point form ; 

(d) y ^mx-{-by slope-intercept form ; 

(e) X cos a + y sin a = p, normal form ; 

(f ) ^x + By + C7 = 0, general form. 

66. tan = — ^ ~ , angle between two lines whose slopes are mi and m^. 
1 + mima 

mi = ms when lines are parallel, 

and mi = when lines are perpendicular. 

ms 

68. (x — a)^ + (y — fi)* = r^) equation of circle with center (a, /9) and radius r. 



4 



DIFFERENTIAL CALCULUS 



2. Greek alphabet. 



Letters 


Names 


A a 


Alpha 


B iS 


Beta 


r 7 


Gamma 


A S 


Delta 


E e 


Epsilon 


z ? 


Zeta 


H «7 


Eta 


e <? 


Theta 


I ( 


Iota 


K K 


Kappa 


A X 


• Lambda 


M It 


Mu 



Letters 


Names 


N V 


Nu 


H f 


Xi 


O 


Omicron 


n IT 


Pi 


P p 


Rho 


2 «r 9 


Sigma 


T T 


Tau 


T V 


Upsilon 


^ 4> 


Phi 


X X 


Chi 


•9 ^ 


Psi 


il <o 


Omega 



3. Natural values of trigonometric functions. 



Angle in 
BadJana 


Angle in 
Degrees 


Sin 


• 

Cob 


Tan 


Got 






.0000 
.0873 
.1746 
.2618 
.3491 
.4363 
.5236 
.0109 
.6981 
.7854 


0« 
5° 
lO'' 
15° 
20° 
25'' 
SOP 
36° 
40° 
45° 


.0000 
.0872 
.1736 
.2588 
.3420 
.4226 
.5000 
.5736 
.6428 
.7071 


1.0000 
.9962 
.9848 
.9659 
.9397 
.9063 
.8660 
.8192 
.7660 
.7071 


.0000 
.0875 
.1763 
.2679 
.3640 
.4663 
.6774 
.7002 
.8391 
1.0000 


GO 

11.430 
5.071 
3.732 
2.747 
2.145 
1.732 
1.428 
1.192 
1.000 


90° 
85° 
80° 
75° 
70° 
65° 
60° 
55° 
50° 
45° 


1.5708 

1.4835 

1.3963 

1.3090 

1.2217 

1.1345 

1.0472 

.9599 

.8727 

.7854 






Cos 


Sin 


Cot 


Tan 


Angle In 
Degrees 


Angle in 
Radians 



COLLECTION OF FORMULAS 



Angle in 
Badians 


Angle in 
Degrees 


Sin 


Cot 


Tan 


Cot 


Seo 


Cm 





(y> 





1 





00 


1 


00 


2 


9(y> 


1 





00 





00 


1 


ic 


180° 





-1 





00 


-1 


00 


Sir 
2 


27(y> 


-1 





00 





00 


-1 


lie 


36(y> 





1 





00 


1 


00 



Angle in 


Angle in 
Degrees 


Sin 


Cos 


Tan 


Cot 


Sec 


Csc 





0« 





1 





00 


1 


00 


fC 
6 


30«> 


1 
2 


2 


s 


V3 


2\/8 
3 


2 


7C 

4 


45« 


2 


V2 
2 


1 


1 


^2 


V2 


3 


60° 


V3 
2 


1 
2 


Va . 


V3 
3 


2 


2V3 
3 


n 
2 


90° 


1 





00 





00 


1 



DIFFEKENTIAL CALCULUS 



4. Rules for signs. 



Quadrant 


Sin 


Cm 


Tan 


Got 


Seo 


G»e 


f irst . . « • 


+ 


+ 


+ 


+ 


+ 


+ 


Second. . . . 


+ 


— 


— 


— 


— 


+ 


Third .... 


— 


— 


+ 


+ 


— 


— 


Fourth. . . . 


— 


+ 


— 


— 


+ 


— 



CHAPTER II 
NUMBERS 

5. Rational numbers. All positive and negative integers and 
fractions, and zero, are called rational numbers. We shall assume 
that the student is familiar with the most elementary properties 
of these numbers and their use in ordinary Arithmetical work. 

6. Comparison of rational numbers with the points of a straight 
line. The series of rational numbers is unlimited, for between 
any two we can always insert as many more rational numbers as 
we please. Nevertheless there exist gaps everywhere in the series, 
as may be clearly seen if we set up a correspondence between the 
series of rational numbers and the points of a straight line. 



o a p 

On a straight line of indefinite length select a zero point and 
a definite unit of length for measuring segments. A length may 
then be constructed corresponding to any rational number a which 
we lay off to the right or left of according as a is positive or 
negative. In this way we obtain a definite end point P which may 
be considered as the point corresponding to the rational number a* 
We may then say, to every rational number there corresponds one 
and only one point on the straight line. 

But there are lengths which are incommensurable with a given 
unit of length. From Geometry we have the familiar example of 
the diagonal of a square whose side is the unit of length. Laying 
off such a length from the origin on the straight line we obtain 
an end point which corresponds to no rational number, f And 

* In aboTO flgnre a to taken as poeitiTe. 

t Length of diagonal of a unit square « v2. This cannot be an Integer, for no integer multi- 
plied by itself gires 2. Neither can it be a fraction ; for, if possible, let 

>^-?. (A) 

where a and h are integers which do not hare a common factor. Squaring both sides, 

a* 
2-_. (B) 

Since a and h hare no common factor, a* and b* can hare no oonmion factor ; and (B), which 
says that 6> is contained twice in a*, contradicts our hypothesis (A). Therefore, since V2 !■ 
neither an integer nor a fraction it cannot be a rational number. 

7 



8 DIFFERENTIAL CALCULUS 

since there are infinitely many lengths which ai*e incommensurable 
with the unit of length, the straight line is infinitely richer in 
point-individuals than the series of rational numbers is in number- 
individuals. This comparison has led to the recognition of a certain 
incompleteness of the system of rational numbers, while we ascribe to 
the straight line completeness and absence of gaps^ that is, continuity, 

7. Irrational numbers. If we wish to study the straight line 
arithmetically, the system of rational numbers having been found 
wanting, it becomes necessary to extend our system of numbers 
in such a way that it shall have the same completeness^ or continuity^ 
as the straight line. This has been done by the creation of irra- 
tional numbers which are defined in terms of rational numbers only. 
The scope of this book does not permit the development of the 
modem arithmetic theory of rational numbers; hence we shall 
only call the attention of the student to the existence of irrational 
numbers and to the statement : the irrational numbers completely 
fill up all the gaps which exist in the system of rational numbers; 
i.e. we assume that to every point on a straight line corresponds 
a number^ rational or irrational^ and conversely. Following are 
examples of irrational numbers : 

V2 = 1.4142136 ..,* 
logioS = 0.6989700 • • -,1 
TT = 3.1415929 •., 
« = 2.7182814... 

8. Real numbers. All rational and irrational numbers are called 
real numbers. These are arranged in order with respect to their 
magnitudes as follows: 

increasing as we pass from left to right. 

• It was shown in footnote on p. 7 that Vi cannot be a rational nomber. 
t Suppose this to be a rational number, then 

h 
where a and ( are positlre integers. Then 

a 

10»«6, or 10" -6». 

That is, no matter what the ralues of a and 6, we would have a number whose last digit is 
zero equal to a number whose last digit is 5 ; this being absurd, our hypothesis that logjoS was a 
rational number is absurd. 



NUMBERS 9 

The symbol > is read is greater than^ and the symbol < is read 
u less than. 

It is sometimes convenient to write a > 0, which means that a 
is positive ; or 6 < 0, showing that h is negative. 

We may also write such expressions as 

3 > - 1, or - 8 < - 5, etc. 

The symbol > is read is greater than or equal to^ and is equiva- 
lent to the symbol <, read is not less than. 

The symbol < is read is less than or equal to^ and is equivalent 
to the symbol >, read is not greater than. 

The symbol ^ is read is greater or less than^ and is equivalent to 
the symbol ¥=, read is not equal to. 

9. Numerical or absolute value. By the numerical value or abso- 
lute value of a real number we mean its value taken positively. 
The numerical or absolute value of a is denoted by the symbol |a|. 

Thus, |5| = I- 51=4-5. 

10. Imaginary numbers. Consider the equation 

a:2 -h 1 = 0. 

No real number substituted for x will satisfy this equation. 
To overcome this difficulty, our number system must be enlarged 
by the creation of a new number. If i is a number such that 
i* = — 1, then the above equation is satisfied by substituting i or 

— t for X. Hence , . 

x=^±% 

is called the solution of the equation, and the new number 
t = V— 1 is termed the imaginary unit. 
If a is real, the expression 

a V— 1, or at, 

defines an imaginary number. 
IL Complex numbers. The sum 

a-\-bij 

where a and b are real numbers, defines a complex number. The 
first term belongs to the system of real numbers, while the second 
belongs to the system of imaginary numbers. Complex numbers 
suffice for all algebraic operations. 



10 DIFFERENTIAL CALCULUS 

12. Division by zero excluded. - is indeterminate. For, the 

quotient of two numbers is that number which multiplied by the 
divisor will give the dividend. But any number whatever multi- 
plied by zero gives zero, and the quotient is indeterminate ; that 
is, any number whatever may be considered as the quotient, a 
result which is of no value. 

- has no meaning, a being different from zero, for there exists 

no number such that if it be multiplied by zero the product would 
equal a. 

Therefore division by zero is not an admissible operation, 

13. Only real numbers considered. Unless otherwise stated, only 
real numbers are considered in what follows in this book. 



CHAPTER III 
VARIABLES AND FUNCTIONS 

14. Variables. A variable is a quantity to which an unlimited 
number of values can be assigned. Variables are denoted by the 
later letters of the alphabet. Thus, in the equation of a straight 

line, 

2 + ^ = 1 
a 

X and y may be considered as the variable coordinates of a point 
moving along the line. 

15. Constants. A quantity whose value remains unchanged is 
called a aynstant. 

Numerical or absolute constants retain the same values in all 
problems, as 2, 5, Vt, tt, etc. 

Arbitrary constants^ or parameters^ are constants to which any 
one of an unlimited set of numerical values may be assigned, 
and they are supposed to have these assigned values throughout 
the investigation. They are usually denoted by the earlier letters 
of the alphabet. Thus, for every pair of values arbitrarily assigned 
to a and 5, the equation 

a 

represents some particular straight line. 

16. Interval of a variable. Very often we confine ourselves to 
a portion only of the number system. For example, we may 
restrict our variable so that it shall take on only such values as 
lie between a and 6, where a and b may be included, or either or 
both excluded. We shall employ the symbol [a, 6], a being less 
than 6, to represent the numbers a, 6, and all the numbers between 
them, unless otherwise stated. This symbol [a, b] is read the 
interval from a to b. 

11 



12 DIFFERENTIAL CALCULUS 

17. Continuous variation. A variable x is said to vary continu- 
ously through an interval [a, 6], when x starts with the value a 
and increases until it takes on the value b in such a manner as 
to assume the value of every number between a and h in the 
order of their magnitudes. This may be illustrated geometrically 

as follows : 

a X b 

O ^ P B 

The origin being at 0, lay off on the straight line the points A 
and B corresponding to the numbers a and b. Also let the point 
F correspond to a particular value of the variable x. Evidently 
the interval [a, b] is represented by the segment AB, Now as x 
varies continuously from a to 6 inclusive, i.e. through the interval 
[a, 6], the point F generates the segment AB. 

18. Functions. When two variables are so related that the 
value of the first variable depends on the value of the second 
variable, then the first variable is said to be a function of the 
second variable. 

Nearly all scientific problems deal with quantities and relations 
of this sort, and in the experiences of everyday life we are con- 
tinually meeting conditions illustrating the dependence of one 
quantity on another. For instance, the weight a man is able to lift 
depends on his strength, other things being equal. Similarly, the 
distance a boy can run may be considered as depending on the time. 
Or, we may say that the area of a square is a function of the length 
of a side, and the volume of a sphere is a function of its diameter, 

19. Independent and dependent variables. The second variable, 
to which values may be assigned at pleasure within limits depend- 
ing on the particular problem, is called the independent variable, or 
argument; and the first variable, whose value is determined as 
soon as the value of the independent variable is fixed, is called 
the dependent variable, or function. 

Frequently, when we are considering two related variables, it 
is in our power to fix upon whichever we please as the fwrfe- 
pendent variable ; but having once made the choice, no change of 
independent variable is allowed without certain precautions and 
transformations. 



VARIABLES AND FUNCTIONS 18 

One quantity (the dependent variable) may be a function of two 
or more other quantities. (the independent variables, or arguments). 
For example, the cost of cloth is a function of both the quality 
and quantity ; the area of a triangle is a function of the hose and 
altitude; the volume of a rectangular parallelopiped is a function 
of its three dimensions. 

20. Notation of functions. The symbol f(x) is used to denote a 
function of x^ and is Tesid f function of x. In order to distinguish 
between different functions the prefixed letter is changed, as 
F{x),4>{x),f(x),eUi. 

During any investigation the same functional symbol always 
indicates the same law of dependence of the function upon the 
variable. In the simpler cases, this law takes the form of a series 
of analytical operations upon that variable. Hence, in such a 
case, the same functional symbol will indicate the same operations 
or series of operations, even though applied to different quantities. 
Thus, if 

f(x) = a? - 9 z 4- 14, 

then /(y) = y'-9y4-14. 

Also f{a) = a* - 9 a 4- 14, 

/(6 4.1) = (6 4- 1)2 - 9(6 4- 1) + 14 = 6» - 76 4- 6, 
/(0)=0»-90 4-l4 = 14, 
/(-l) = (-l)»-9(-l)4- 14 = 24, 
/(3)=3*-9-3+14=-4, 
/(7)=7*-9-7 4-14 = 0, etc. 

Similarly ^(Xy y) denotes a function of x and y, and is read 
<t> function of x and y. 

If 4>(x,y) = 8in(x-{-y)y 

then <t> (a, b) = sin (a 4- 6), 

and *^ ( "o ' ) = si^i o ~ ^• 

Again, if F(xy y, z) = 2 a? 4- 3 y — 12 «, 

then F{m, — tw, w) = 2 m — 3 w — 12 m = — 13 wi, 

and F{S, 2, 1)= 2 3 4- 32 - 12 1 = 0. 

Evidently this system of notation may be extended indefinitely. 



14 DIFFERENTIAL CALCULUS 

21. Values of the independent variable for which a function is 
defined. Consider the functions 

a:* — 2 a: H- 5, sin a;, arc tan x 

of the independent variable z. Denoting the dependent variable 
in each case by y, we may write 

y = a? — 2a;4-5, y = sina:, y = arc tan x* 

In each case y (the value of the function) is known, or, as we 
say, defined^ for all values of x. This is not by any means true of 
all functions, as the following examples illustrating the more com- 
mon exceptions will show. 

Here the value of y (i.e. the function) is defined for all values 
of X except a; = 6. When a: = 6 the divisor becomes zero and the 
value of y cannot be computed from (1). Any value might be 
assigned to the function for this value of the argument. 

(2) y = V;. 

In this case the function is defined only for positive values of x. 
Negative values of x give imaginary values for y, and these must 
be excluded here where we are confining ourselves to real num- 
bers only. 

(3) y = log„a:. a > 

Here y is defined only for positive values of x. For negfative 
values of x this function does not exist (see § 85). 

(4) y = arc sin x^ y = arc cos x. 

Since sines and cosines cannot become greater than -h 1 nor less 
than — 1, it follows that the above functions are defined for all 
values of x ranging from — 1 to -f 1 inclusive, but for no other 
values. 

22. One-valued and many-valued functions. A variable y is said 
to be a one-valued function of a second variable x when y has one 
and only one value corresponding to each value of a?. Thus, in 

y = 3a? 

y is a one-valued function of x. 



VARIABLES AND FUNCTIONS ' 15 

If to each value of the second variable there correspond more 
than one value of the first variable, then the first variable is said 
to be a many-valued function of the second variable. In 

y is a two-valued function of x since 

Again, in y = arc tan x 

it is seen that there is no limit to the number of values of y corre- 
sponding to a given value of x. For, let a? = 0, then y = wtt, where 
n denotes zero or any integer. 

23. Explicit functions. When a relation between x and y is 
given by means of an equation solved for y, then y is called an 
explicit function of x. Thus, in 

-y/x — 3 
y = sin aa:, y = log(l 4- x), y = 5 a*, 

y is in each case an explicit function of x. 

Again, in 2 = log {x -f y) 

2 is an explicit function of x and y. 

Similarly w^e'^ 

exhibits t£f as an explicit function of x^ y, and z. 

Symbolically these explicit functions may be respectively denoted 

z = <^(a;, y), 
w = F(x, y, z). 

24. Inverse functions. Let y be given as a function of x by 
means of the relation 

It is usually possible in the case of functions considered in this 
book to solve this equation for x, giving 

a; = <^(y); 



18 DIFFERENTIAL CALCULUS 

(4) Inverse trigonometric functions ; as arc sin a:,* arc cot (a; — y), 
etc. 

Many more transcendental functions are studied in the higher 
branches of mathematics. 

EXAMPLES 

1. GiYen f{x) = x» - 10 x* 4- 81 x - 30 ; show that 

/(O) = - 30, f(y) = y» - 10y2 + Sly - 30, 

/(2) = 0, f(a) = a« - 10a2 -f 31 a - 30, 

/(3) = /(6),' f(yz) = y»«» ~ 10 yH^ + 31 yz - 30, 

/(I) >/(- 3), /(x - 2) = x» - 16x2 + 83x - 140, 
/(-I) = -6/(6). 

2. If/(x) = x'-10x« + 31x-30, and (x) = X* - 66x2 ~210x- 216; showihat 

/(2) = 0(-2), 
/(3) = 0(-3), 
/(5) = 0(-4), 
/(0) + 0(0) + 246 = 0. 

8. GiYen F(x) = x(x - 1) (x 4- 6) (x - J) (x + |) ; show that 

F(0) = F{1) = F(- 6) = F(i) = F{- i) = 0. 

m — 1 

4. If /(wii) = — ; show that 

mi 4- 1 

/('Wi) —f(m2) mi — mt 
1 + /(wii)/(m2) 1 4- w»im« 

1 -x 

5. Given 0(x) = log ; show that 

1 4-x 

0(x) + 0(y) = 0(^^). 

6. If /(0) = cos ; show that 

/W =/(- 0) =-/(^ - 0) = -/(«^ + 0). 

7. If F(0) = tan ; show that 

^ ' i-[2^(W' 

8. Given ^(x) = x'" 4- x**" 4- 1 ; show that 

^(1) = 3, \^(0) = 1, ^(a) = ^(-a). 

*A1bo written sin-ix, the -1 not being considered as a negative exponent in the ordinary 

sense, but merely indicating the inverse function. The expression y = arc sin x should be read 

y equalt the arc (or angle) whose »ine it Xf and the same relation between x and y is given by 

sin y^x. 

For example, since tan - = 1 , 

4 

we may also write - «> arc tan 1. 

4 



CHAPTER IV 

THEORY OF LIMITS 

29. Limit of a variable. If a variable v takes on successively a 
series of values that approach nearer and nearer to a constant 
value I in such a manner that \v — l\* becomes and remains less 
than any assigned arbitrarily small positive quantity, then v is said 
to approach the limit l^ or to converge to the limit l. Symbolically 
this is written ^j^i^ v = L 

The following familiar examples illustrate what is meant. 

(1) As the number of sides of a regular inscribed polygon is in- 
definitely increased, the limit of the ai*ea of the polygon is the ai*ea 
of the circle. In this case the variable is always less than its limit 

(2) Similarly the limit of the area of the circumscribed poly- 
gon is also the area of the circle, but now the variable is always 
greater than its limit. 

(8) Consider the series 

{A) i_^ + j_j+.... 

The sum of any even number (2n) of the first terms of this 
series is 1 i 1 i i 

*^2ii ■■■ 2 4 8 2*""' 2**~*' 

,Bs s -— -? ^ By6,p.l 

(20 ^..-_^_i-3 3.2..-I- 

Similarly the sum of any odd number (2 n + 1) of the first 
terms of the series is 

"fn+i ■*■ 2 4 8 ' 2*""' 2*"' 
1 



1 



((h S -_?!ll!—-2, 1 By6,p.l 

* To be read the numerical value qf the difference between v and L 

10 



20 DIFFERENTIAL CALCULUS 

Writing (B) and (O) in the forms 

^ u « limit /2 o ^— limit ^ a 

we have ^ I o "" ^a« ) = q ni»^i ^ 0, 

and n = aoV^»-^^"3;-n = oo3T2^--®- 

Hence by definition of the limit of a variable it is seen that both 
S^^ and iS^s.^i are variables approaching § as a limit as the num- 
ber of terms increases without limit. 

Summing up the first two, three, four, etc., terms of {A)j the 
sums are found by (B) and ((7) to be alternately less and greater 
than ), illustrating the case when the variable^ in this case the sum 
of the terms of (il), i» sometimes less and sometimes greater than 
its limit. 

In the examples shown the variable never reaches its limit This 
is not by any means always the case, for from the definition of 
the limit of a variable it is clear that the essence of the definition 
is simply that the numerical value of the difference between the 
variable and its limit shall ultimately become and remain less than 
any positive number we may choose however small. 

(4) As an example illustrating the fact that the variable may 
reach its limit, consider the following. Let a series of regular 
polygons be inscribed in a circle, the number of sides increasing 
indefinitely. Choosing any one of these, construct the circum- 
scribed polygon whose sides touch the circle at the vertices of 
the inscribed polygon. Let p^ and F^ be the perimeters of the 
inscribed and circumscribed polygons of n sides and C the circum- 
ference of the circle, and suppose the values of a variable a; to be 
as follows : 

Then evidently, 

Tl = OO ' 

and the limit is reached by the variable. 



THEORY OF LIMITS 21 

30. Infinitesimals. A variable v whose limit is zero is called 
an infinitenmcd.* This is written 

limit v = 0, 

and means that the successive numerical values of v ultimately 
become and remain less than any positive quantity however small. 
Such a variable is said to become indefinitely small or to idtimatelt/ 
vanish. 

If limit V = Z, then limit (v — Z) = ; 

that is, the difference between a variable and its limit is an infini- 
tesimal. 

Conversely, if the difference between a variable and a constant is 
an infinitesimal^ then the variable approaches the constant as a 
limit. 

31. The concept of infinity (oo). If a variable v ultimately 
becomes and remains greater than any assigned positive number 
however large, we say v increases without limits and write 

limit t; = 4- 00. 

If a variable v ultimately becomes and remains algebraically 

less than any assigned negative number, we say v decreases without 

limity and write 

limit t; = — 00. 

If a variable v ultimately becomes and remains in numerical 
value greater than any assigned positive number however large, 
we say v, in numerical value^ increases without limits or v becomes 
infinitely great^jf and write 

limit v = 00. 

Infinity (oo) is not a number; it simply serves to characterize 
a particular mode of variation of a variable by virtue of which it 
increases or decreases without limit. 

* Hence a constant, no matter how small it may be, is not an InfinitMimal. 

t On account of the notation used and for the sake of uniformity, the expression limit v « -I- « is 
sometimes read v approachet the limit plua infinity. Similarly limit r » - « is read v approaehea 
the limit minus infinity , and limit t; » « is read v, in numericcU value, approaches the limit 
it^nity. 

While the abore notation is conrenient to use in this connection, the student must not forget 
that infinity is not a limit in the sense in which we defined a limit on page 19, for infinity is not 
a number at all. 



22 DIFFERENTIAL CALCULUS 

32. Limiting value of a function. Given a function f{x). 

If the independent variable x takes on any series of values 

such that 

limit 2; = a, 

and at the same time the dependent variable /(2;) takes on a series 
of corresponding values such that 

limit /(a:) = A^ 

then as a single statement this is written 

and is read the limit of f{x)^ as x approaches the limit a hy any set 
of values^ is A.* 

33. Continuous and discontinuous functions. A function f{x) is 
said to be continuous for x = aii the limiting value of the function 
when X approaches the limit a in any manner is the value assigned 
to the function for 2; = a. In symbols, if 

then f{x) is continuou4i for x==a. 

The function is said to be discontinuous for x = a if this con- 
dition is not satisfied. For example, if 

limit J*/ V 

the function is discontinuous for x=a. 

The attention of the student is now called to the following cases 
which occur frequently. 

Case I. As an example illustrating a simple case of a func- 
tion continuous for a particular value of the variable, consider 
the function 

For a:= 1, /(a:)=/(l)= 3. Moreover, if x approaches the limit 
1 in any manner, the function f(x) approaches 3 as a Umit. Hence 
the function is continuous for x = l. 

* It Bometimes happens that/Cz) approaches one limit when x approaches a, x being always 
less tlian a ; and a different limit when x approaches a, x being always greater than a. 0r,/(x) 
may approach a limit from one side and not from the other ; or it may approach no limit from 
either side. Evidently the above definition excludes all snch exceptional cases. 



THEORY OF LIMITS 23 

Case II. The definition of a continuous function assumes that 
the function is already defined for 2; = a. If this is not the case, 
however, it is sometimes possible to assign such a value to the 
function for x = a that the condition of continuity shall be satis- 
fied. The following theorem covers these cases : 

Theorem. Iff{x) is not defined for a; = a, and \f 

then f(x) will be continuous for x=^a^if B is assumed as the value 
off(x) for x^a. Thus the function 

x-2 

is not defined for x^2 (since then there would be division by 
zero). But for every other value of re, 

and 2; _ 2 (^ "^ ^^ ~ * * 

. , . limit re* — 4 

therefore o K = 4. 

x= A X — 2 

Although the function is not defined for rr = 2, if we arT)itrarily 
assign it the value 4 for rr = 2, it then becomes continuous for this 
value. 

A function f{x) is said to he continuous in an interval when it is 
continuous for all values of x in this interval,* 

34. Continuity and discontinuity of functions illustrated by their 
graphs. 

(1) Consider the function rr*, and let 

{A) y = 2^. 

* In thfs book we shall deal only with f onctlons which are In general oonttnnons, that !b, con- 
tlnnona for all ralues of x, with the possible exception of certain isolated yalues, our resalts in 
general being understood as yalid only for snch ralaes of x for which the function in question 
is actually continuous. Unless special attention is called thereto, we shall as a rule pay no 
attention to the possibilities of such exceptional yalues of x for which the function is discon' 
tinuous. The definition of a continuous function /(x) is sometime roughly (but imperfectly) 
summed up in the statement that a small change in x thail produce a tmall change in /(x). We 
shall not consider functions baring an infinite number of oscillations in a limited region. 



24 



DIFFEKENTIAL CALCULUS 




If we assume values for x and calculate the corresponding values 
of y^ we can plot a series of points. Drawing a smooth line free- 
hand through these points a good representation 
of the general behavior of the function may be 
obtained. This picture or image of the function 
is called its graph. It is evidently the locus of all 
points satisfying equation {A). 

Such a series or assemblage of points is also 
^ called a curve. Evidently we may assume values 
of X so near together as to bring the values of y (and therefore 
the points of the curve) as near together as we please. In other 
words, there are no breaks in the curve, and the function ^ is 
continuous for all values of x. 

(2) The graph of the continuous function sinx is plotted by 
drawing the locus of 

y = sin X. 

It is seen that no break in the 
curve occurs anywhere. 





(3) The continuous function «* 

quent occurrence in the Calculus. 

graph from 

y = e» 

we get a smooth curve as shown. 
' is clearly seen that, 



is of very fre- 
If we plot its 

(« = 2.718 ••) 
From this it 



(a) when a;== 0, ^"^^^ y( = e^ = 1; 

(b) when a? > 0, y(= e') is positive and increases as we pass 
towards the right from the origin; 

(c) when x<0, y (= e*) is still positive and decreases as we 
pass towards the left from the origin. 

(4) The function log^x is closely related to 
the last one discussed. In fact, if we plot its 
graph from 

y = log.ar, 

it will be seen that its graph has the same 

relation to OX and OF as the graph of e* has to OF and OX. 




THEORY OF LIMITS 



25 



Here we see the following facts pictured : 

(a) Fora: = l, log.a; = log.l = 0. 

(b) For 2; > 1, log^x is positive and increases as x increases. 

(c) For 1 > a; > 0, log;a; is negative and increases in nwmerical 
value as x diminishes. 

(d) For 2; < 0, log^a; is not defined ; hence the entire graph lies 
to the right of OY. 

(5) Consider the function -9 and set 

X 

1 

y = — 

X 

If the graph of this function be plotted, it will be seen that as x 
approaches the value zero from the left (neg* 
atively) the points of the curve ultimately 
drop down an infinitely great distance, and 
as X approaches the value zero from the right 
the curve extends upward infinitely far. 

The curve then does not form a continuous 
branch from one side to the other of the axis 
of F, showing graphically that the function 
is discontinuous for 2; = 0, but continuous for all other values of x, 

(6) From the graph of 



o 




y = 



2x 



it is seen that the function 

2a? 

is discontinuous for the two values a; = ± 1, but continuous for all 
other values of x. 
(I) The graph of 

y = tana; 

shows that the function tan z is 
discontinuous for infinitely many 

values of ar, namely, a; = — , where 

n denotes any odd positive or negative integer. 



I 







26 



DIFFERENTIAL CALCULUS 



(8) The function 

arc tanrr 

has infinitely many values for a given value of Xj the graph of 




equation 



y = arc tan x 



consisting of infinitely many branches. If, how- 
ever, we confine ourselves to any single branch, 
^ the function is continuous. For instance, if we 
say that y shall be the arc of smallest absolute 
value whose tangent is 2;, that is, y shall take 

on only values between — — and — » then we are limited to the 

branch passing through the origin and the condition for continuity 
is satisfied. 



(9) Similarly 



arc tan - 

X 



is found to be a many-valued function, 
branch of the graph of 



Confining ourselves to one 



y = arc tan -» 

X 

we see that as x approaches zero from 
the left y approaches the limit — — , and 




TT 



as X approaches zero from the right y approaches the limit + •— 

Hence the function is discontinuous when a; = 0. Its value for 
2; = can be assigned at pleasure. 

Functions exist which are discontinuous for every value of the 
independent variable within a certain range. In the ordinary ap- 
plications of the Calculus, however, we deal with functions which 
are discontinuous (if at all) only for certain isolated values of 
the independent variable ; such functions are therefore in general 
continuous, and are the only ones considered in this book. 

35. Fundamental theorems on limits. In problems involving 
limits the use of one or more of the following theorems is usually 
implied. It is assumed that the limit of each variable exists and 
is finite. 



THEORY OF LIMITS 27 

Theorem I. The limit of the algebraic sum of a finite number of 
variables is equal to the like algebraic sum of the limits of the several 
variables. 

Theorem II. The limit of the product of a finite number of vari- 
ables is equal to the prodvxA of the limits of the several variables. 

Theorem III. The limit of the quotient of two variables is equal 
to the quotient of the limits of the separate variables^ provided the 
limit of the denominator is not zero. 

Before proving these theorems it is necessary to establish the 
following properties of infinitesimals. 

(1) The sum of a finite number of infinitesimals is an infinitesimal. 
To prove this we must show that the numerical value of this sum 
can be made less than any small positive quantity (as e) that may 
be assigned (§ 30). That this is possible is evident, for, the limit 
of each infinitesimal being zero, each one can be made numerically 

less than - (n being the number of infinitesimals) and therefore 

their sum can be made numerically less than e. 

(2) The product of a constant c and an infinitesimal is an infinir 
tesimal. For the numerical value of the product can always be 
made less than any small positive quantity (as e) by making the 

numerical value of the infinitesimal less thaif -• 

c 

(3) The product of any finite number of infinitesimals is an infini- 
tesimal. For the numerical value of the product may be made 
less than any small positive quantity that can be assigned. If the 
given product contains n factors, then since each infinitesimal may 
be assumed less than the nth root of e, the product can be made 
less than e itself. 

(4) If V is a variable which approaches a limit I different from 
zero^ then the quotient of an infinitesimal by v is also an infinitesi- 
mal. For if limit v = 1, and k is any number numerically less 
than {, then by definition of a limit, v will ultimately become and 

remain numerically greater than k. Hence the quotient -i where 

€ is an infinitesimal, will ultimately become and remain numerically 

less than -^ and is therefore by (2) an infinitesimal. 



28 DIFFERENTIAL CALCULUS 

Proof of Theorem L Let v^^ v^ v„ • • • be the variables, and i^^ 
l^ !,, • • • their respective limits. We may then write 



where 6^ e,, Cg, • • • are infinitesimals (i.e. variables having zero for 
a limit). Adding, 

(A) (Vj + r, + V, H ) — (^ + ^1 + ^1 H ) = («! + €t + €g H ). 

Since the right-hand member is an infinitesimal by (1), p. 27, 
we have from the converse theorem on p. 21, 

limit (Vj + V, + V, H ) = ?i + ?, + ?8 "• 

or, limit (v^ + v, + v, H ) = limit v^ + limit v, + limit v, H , 

which was to be proved. 

Proof of Theorem U. Let v^ and r, be the variables, l^ and 
/, their respective limits, and c^ and e, infinitesimals; then 

and r^ = ?, + €,. 

ft 

Multiplying, v^v, = {l^ + c^) (Z, -h €,) 

or. 

Since the right-hand member is an infinitesimal by (1) and (2), 
p. 27, we have as before 

limit (VjVj) = Z^Zj = limit Vj • limit v,, 

which was to be proved. 

Proof of Theorem lU. Using same notation as before. 



Vg Zg-h€, Z, \Z« + c, V 



or 



9 






THEORY OF LIMITS 



29 



Here again the right-hand member is an infinitesimal by (4), 
p. 27, if Z, =#= 0, hence 

limit r^V^ = ^°'^^^S 
\v^J i, limit V, 

which was to be proved. 

It is evident that if any of the variables be replaced by con- 
stants our reasoning still holds and the above theorems are true. 

36. Special limiting values. The following examples are of 
special importance in the study of the Calculus. In the first 
twelve examples a> and c =#= 0. 



Written in the farm of limits. 



Abbreviated form often used. 



(1) 
(2) 
(8) 



limit £ ^ 



« = 



oo: 



X 



limit 



ca; = oo; 



c 

— = 00. 





<? . 00 = 00. 



limit £ _ 



limit ^ 
« = QOa. 

a* 
a' 
a* 
a* 



oo; 



-=0; 



(4) 

(5) 
(6) 
(7) 
(8) 
(9) X S'o ^°Sa a: = 4- 00, when a < 1 

(10) li"^* log„ a; = - 00, when a < 1 

(11 ) "^'o ^^?« a: = - 00, when a > 1 

(12) x='i«^^»<»^ = + ^' whena>l 



00 

c 
c 

00 



— = 00. 



-^ = 0. 



limit 

aj =— oo 

limit 

X = + CO 

limit 
« = — « 

Umit 

a; = + CO 



= -f 00, when a < 1 
= 0, when a < 1 
= 0, when a > 1 
= -f 00, when a > 1 



a""" = 4-00. 



a+« = 0. 
a— = 0. 



a+* = -f 00, 



lOga = + 00. 

log. (+ oo) = - 00. 

loga = - 00. 
^^ga{-h QO) = + 00. 



The expressions in the second column are not to be considered 
as expressing numerical equalities (oo not being a number) ; they 
are merely symbolical equations implying the relations indicated in 
the first column, and should be so understood. 



80 DIFFERENTIAL CALCULUS 

limi^ !l* — fl.* 

(13) Find _ , where n denotes any positive integer.* 

By division we get 

^-^ = af -* + 02^-* -h 0*2*"' -h • • • + a*"'a; + a""^ 
X — a 

for every value of x except x = a. Therefore 

limit ^-ar ^ limit ^-i limit ^., ^ _^ limit ^,,,^ limit ^..,^ 
x = a x — a x = a x = a x = a x=^a 

[By Theorem I, p. 27.] 

The limit of each term in the second member is a"~^; and since 
there are n terms, we have 

limit af - a* ^_, 
X = a X — a 

limit sin a; 



(14) Show that "_ ^ 



= L 



Let be the center of a circle w^hose radius is unity. 
Let arc-4Jlf = arc-4Jlf' = a;, and let MT and M^T be tangents 
drawn to the circle at M and M\ From Geometry, 

MPM' < MAM' < MTW\ 
2 sin a; < 2 a; < 2 tan z. 

Dividing through by 2 sin a: we get 

^ X 1 
1 <-T — < 




sm X cos X 
If now X approaches the limit zero, 

limit X 



a: = si 



sma; 



lllYllf 1 

must lie between the constant 1 and ^ , which is also 1. 

X = \) cos X 

rnu f ^ limit ^ i ^« limit sin a; ^ ^, ^^^ ^- 

Therefore ^ -: — = 1, or ^ = 1. Th. Ill, p. 27 



a: = sin a; ' a; = 



* Restricted to a positive integer in order to simplify the work. The result holds true for all 
values of n. 



THEORY OF LlMlTS 



81 



It is interesting to note the behavior of this function from its 
graph, the locus of equation 

sin a; 



X 




-3«- -1 



Although the function is not defined for 2; = 0, yet it is not 
cdscontinuous when 2; = if we define 

sin 







= 1. 



Case II, p. 23 



(15) Find the limit of the sum of the series 

.11 1 

1 H 1 1 1 » 

3 9 3"-^ 

as the number of terms increases without limit. 

By formula 6, p. 1, we find that the sum of n terms of the 
series is 



Hence 



limit ^ 
n = 00 



Umit 3 
w = 00 2 



^ 3 limit A _ 1\ 
2 w = 00 \ 3-y 

___3r limit f't\_ limit /JLN"] 
""2Ln = oo^^ w = oo V3-y J 

= |[l-0]-|. An. 



37. The number e.* We first proceed to prove two important 
theorems. 

Theorem I. If a is a variable > — 1 which varies continuously 
in any interval not including zerOj then the function 

• <^ (a) = (1 -h a)« 
varies in a sense contrary to a, 

* The proofs in this section are due to Vall^-Poussin. 



82 DIFFERENTIAL CALCULUS 

We start with the identity 

or, a"+^ = l +(a-l)(a'' + a—^4--- + a + l); 

where a denotes any positive number and n a positive integer. 

Since the last parenthesis is > or < (n + 1) according as a is > 

or < 1, or according as (a — 1) is positive or negative, we have in 

either case , ^ ^ ^ 

a*+^>l4-(n + l)(a~l). 

Let w be any number > — n and different from zero, and 
replace a in this inequality by the quotient 

1 + - J and reducing gives 



^^TTx) >^i 



)■ 



Let m be any integer > n ; then by repeated applications of 
the last result we see that 



('^'^)' >('<)'■■ 



and it readily* follows that 

according as oi < 0. Now replacing <o by ma we get 

(A) a + a)l>(l+^af, 

according as a ^ 0. This proves our statement for two values (as 
a and — a) having the same sign and whose quotient is rational. 

In order to extend this proof to the case of two values a and /3 
whose ratio is irrational, consider first the case when a > 0. 

* The Msumption is here made that raising both members of an inequality to any power, 
rational or irration€U, does not or does change the sense of the Inequality according as the power 
is positire or negatire 



THEORY OF LIMITS 83 

Assuming a < /3 we may so choose m and n that 

n 
Now let — a approach yS as a limit by a series of increasing 

values ; then the second member of the inequality {A) will be 
constantly decreasing and we get in the limit since the function 
is continuous i i 

(1 -h ay > (1 -h M 
or, (f>{a)> <f>{/3). a</3 

Similarly when a < we shall get 

<f>{a)<<l>(/3), a>l3 

which establishes the theorem for all cases. 

Theorem II. Definition of the number e. As a approaches the 
limit zero^ the function 

<^(a) = (l + a)= 

of Theorem I approaches a limit. In whatever manner a approaches 
zerOy the limit is the same number. This limit is denoted by e. 
Consider the two variables a and /3 connected by the condition 

1 + a = r -'i where > a > — 1 ; 

then it follows that 

<^(a) = (l+/8)<^(/8).* 

If a tends towards zero through a series of increasing (therefore 
negative) values, /8 will be positive and tend towards zero through 
a series of decreasing values. Then, by Theorem I, we know that 
<f>{a) continually decreases and ^(^) continually increases. But 
<f>(a) always remains positive and therefore must tend towards 
some definite limit (see Theorem II, above). Denoting this limit 



' -1 



• ♦(a)-(l-».a)«=(l + -i^-l)i-^^ [«-j^-lbyhypotheiiB.] 



1 + 3 1 + 3 



(ji-) ^ =(1 + 3) ^ -(l + 3)(l + 3)^=(l + 3)0(3). 



34 



DIFFERENTIAL CALCULUS 



by e, the last equation, in which 1 + fi tends towards the limit 
unity, becomes in the limit 



limit 



limit 



Since a was negative and /3 positive, this proves that the limit is 
the same whatever may be the sign of the variable. 

To evaluate this limit we note that ^ (ff) increases towards its 
limit, while <f)(a) = {l -{- fi)<f) (^) decreases towards the same limit 
(= e). Hence for all positive values of )S, 

<f>{/3)<e<{l-h^)4>{l3).^ 

By means of this inequality we may calculate the value of « to 
any desired degree of accuracy by choosing /3 sufficiently small. 
If we let )3 = i, then 

(I)' <«<(!)•; 

hence e certainly lies between 2 and 4. In Chapter XX, Ex. 14, 

p. 237, we give a more expeditious method for calculating e. 

Approximately 
^^ ^ € = 2.71828 •••. 

Plotting the graph of <f>(a) from 

y=(l+a)% 

and assigning to y the value e when a = 0, 
we see that as a increases without limit t/ 
approaches the limit 1, and as a approaches 
the limit —1 from the right y increases 
without limit. 
Natural logarithms are those which have the number e for base. 
These logarithms play a very important r81e in mathematics. 
When the base is not indicated explicitly, the base e is always 
understood in what follows in this book. Thus log^v is written 
simply log v. 

Natural logarithms possess the following characteristic prop- 
erty: If a approaches zero as a limit in any way whatever, 

limit ^Qg(^"^^) ^ limit log (1 + a)* = log e = 1. 





^ 


e 


^4 
1 




Ss,^^^ y-e 






1 y-i 


-1 





a 



THEORY OF LIMITS 86 



EXAMPLES 

Prove the following. 



1. ««"'» (±±1) = 1. 

Proof. Un.it/x + lN umit/j 1\ 

limit ..V . limit /1\ m. t «^ 

= X = «<^>+x = ao(i) Th.I,p.27 



[Since these limits exist.] 
= 1+0 = 1. 



2 limit / g'-f 2g \ 1 



Proof, »°^it /?!±2^\ ^ limi 
X = ap\6_3xa/ x = ( 



.» + - 

limit ( as 




[DiYidfiig both namerator and denominator by x>.] 

limit /i . ?\ 
x = ooV ■'"x/ 



limit / 6^ _ 3'\ 
x = «Vx2 / 
[Since the limit of the denominator is not zero.] 

limit .jv , limit /2\ 
X = op^ ' "*■ X = oo\x/ 

limit f^\_ limit .o\ 
X = ao\x2/ x = oo' ^ 

■ 

[Since these limits exist.] 
_ 1 +0 _ 1 
~0-3" 3* 



Th. m, "p. 27 



Th. I, p, 27 



«• x = l^^Tf^ = ^- «• j[»-^J(3ax«-2Ax + 5^^) = 3ax«. 

4 limit 8xM^6^ ^ _ 2 n^it , ^ 5^ ^. ,) = «. 

5 limit ?1±1^5 o limit (g -k)^-2 fcc» _ 
' x = -2 X4-3 • * k = x(x-\-k) 

»• Jj^^Q[2 8in(a + wix)co8{a-mx)] = 8in2a. 

^n limit r • /tt + «\ /u-»\"| 

11. "^" [t«i«(y - ») - cos V] = seca tf. 

If "*" " 



86 DIFFERENTIAL CALCULUS 

. A limit cos (a — a) 

1^- » — -jz ~ = - tan o. 

a = -cos (2 a — a) 

«i« limit /„^.„„ 1 +x\ tiT - . ^^ . 

*'^' ^ _ Q ( arc tan \ = — - , n bemg an odd integer. 

!*• yZ\ <*« COS Vl - yi) =^ , » being an odd integer. 

16. ]'?'J I (ei + e-i) = a. 20. ]"»" ?llli = 8. 

Y^ limit 6x«-2z _ 



x = oo ^ -*• 



22. ^^^^^[cos{^ + ;i)?f|^] = cos(?. 



18. ^*™>t -J^ = 1. oo limit tan 



y = <»y + i \ ''''•0 = 







19. ^^™^^ ri(n^-\) ^ ijjjj.^ 1 - cos ^ 1 

n = oo(n4-2)(n + 3) ' '^^ (? = ^ — = 2* 

OE limit 1 .... 

X = a i^T^ = - 00, If X is increasing as it approaches the value o. 

Oft limit 1 , •, . J 

« = a i^TS = +«♦"« IS decreasmg as it approaches the value o. 



CHAPTER V 

DIFFERENTIATION 

38. Introduction. We shall now proceed to investigate the 
manner in which a function changes in value as the independ- 
ent variable changes. The fundamental problem of the Differ- 
ential Calculus is to establish a measure of this change in the 
function with mathematical precision. It was while investigating 
problems of this sort, dealing with continuously varying quantities, 
that Newton * was led to the discovery of the fundamental prin- 
ciples of the Calculus, the most scientific and powerful tool of the 
modem mathematician. 

39. Increments. The increment of a variable in changing from 
one numerical value to another is the difference found by sub- 
tracting the first value &om the second. An increment of x is 
denoted by the symbol Ax, read delta x. 

The student is warned against reading this symbol '^ delta times 
a:," it having no such meaning. Evidently this increment may be 
either positive or negative f according as the variable in changing 
is increasing or decreasing in value. Similarly 

Ay denotes an increment of y, 
A0 denotes an increment of ^, 
A/^(ar) denotes an increment of /(a;), etc. 

If in y =f{x) the independent variable x takes on an increment 
Ax, then Ay is always understood to denote the corresponding 
increment of the function f(x) (or dependent variable y). 

* Sir Isaac Newton (1642-1727), an Englishman, was a man of the most extraordinary genius. 
He developed the science of the Galculiis under the name of Fluxions. Although Newton had 
dlsooTcred and made use of the new science as early as 1670, his first published work in which it 
oooure Is dated 16S7, having the title Philosophiae Naturalis Principia Mathematica. This was 
Newton's principal work. Laplace said of it, '*It will always remain preeminent above all other 
productions of the human mind." 

t Some writers call a negative increment a decrement, 

37 



88 DIFFERENTIAL CALCULUS 

The increment Ay is always assumed to be reckoned from a defi- 
nite initial value of y corresponding to the arbitrarily fixed initial 
value of X from which the increment A2; is reckoned. For instance, 
consider the function 

Assuming 2: = 10 for the initial value of x fixes y = 100 as the 
initial value of y. 

Suppose X increases to a; = 12, that is, Ax = 2 ; 

then y increases to y = 144, and Ay = 44, 

Suppose X decreases to a: = 9, that is, Aa; = — 1 ; 

then y decreases to y = 81, and Ay = — 19. 

It may happen that as x increases y decreases, or the reverse ; 
in either case A2: and Ay will have opposite signs. 

It is also clear (as illustrated in the above example) that if 

is a continuous function and A2; is decreasing in numerical value, 
then Ay also decreases in numerical value. 

40. Comparison of increments. Consider the function 

{A) y = a?. 

Assuming a fixed initial value for 2;, let x take on an increment 

^x. Then y will take on a corresponding increment Ay, and we 

have 

y -f Ay = (a; -f Aa;)^ 

or, y -f Ay = a^ -f 2 ic • Aa; + (Aa;)*. 

Subtracting {A\ y =x^ 

(B) Ay= 2 a: • Aa; -h (Aa-)* 

we get the increment Ay in terms of x and Aa;. 

To find the ratio of the increments, divide (B) by Aa:, giving 

^ = 2 a; -f Aa;. 
Aa; 

If the initial value of a; is 4, it is evident that 

limit ^^3 
Aa; = Aa; * 



DIFFERENTIATION 



89 



Let us carefully note the behavior of the ratio of the increments 
of X and y as the increment of x diminishes. 



Initial 


New 


Increment 


Initial 


New- 


Increment 


Ay 


value of X 


value of X 


dx 


value of y 


value of y 


Ay 


Ax 


4 


6.0 


1.0 


16 


26. 


9. 


9. 


4 


4.8 


0.8 


16 


23.04 


7.04 


8.8 


4 


4.6 


0.6 


16 


21.16 


6.16 


8.6 


4 


4.4 


0.4 


16 


19.36 


3.36 


8.4 


4 


4.2 


0.2 


16 


17.64 


1.64 


8.2 


4 


4.1 


0.1 


16 


16.81 


0.81 


8.1 


4 


4.01 


0.01 


16 • 


16.0801 


0.0801 


8.01 



It is apparent that as ^x decreases Ly also diminishes, but their 
ratio takes on the successive values 9, 8.8, 8.6, 8.4, 8.2, 8.1, 8.01; 

illustrating the fact that -^ can be brought as near to 8 in value 

as we please by making Ax small enough. Therefore 

limit Ay^g 
Aa; = Aic 

41. Derivative of a function of one variable. The fundamental 
definition of the Differential Calculus is : 

The derivative * of a function is the limit of the ratio of the incre- 
ment of the function to the increment of the independent variable^ when 
the latter increment approaches the limit zero. 

When the limit of this ratio exists, the function is said to be 
differentiable^ or to possess a derivative. 

The above definition may be given in a more compact form 
symbolically as follows: Given 

and assume for x some value for which f{x) is continuous. 

Let X take on an increment Ax ; then y takes on an increment 
Ay, the new value of the function being 

(B) y-^Ay =^f{x -h Ax). 

* Also called the differential coefficient or the derived function. 



40 DIFFERENTIAL CALCULUS 

To find the increment of function, subtract (A) from (5), giving 
(C) Ay=/(a^-f Aa:)-/(a:). 

Dividing by the increment of the variable Ax^ we get 

^ ' Ax Ax 

The limit of this ratio when Ax approaches the limit zero is, 

from our definition, the derivative and is denoted by the symbol -j-* 

dx 

Therefore 

.rrv * dy^ limit /(a? + Ax) - f(x) 

defines the derivative of y \prf(x)] with respect to x. 
From (2>) we also get 

dy _ limit A|^. 

The process of finding the derivative of a function is called 
differentiation. 

It should be carefully noted that the derivative is the limit of 
the ratio, not the ratio of the limits. The latter ratio would assume 

the form -> which is indeterminate (§ 12, p. 10). 

42. Symbols for derivatives. Since Ay and Ax are always finite 
and have definite values, the expression 

Ay 

Ax 

is really a fraction. The symbol 

dy^ 
dx 

however, is to be regarded, not as a fraction, but as the limiting 
value of a fraction. In many cases it will be seen that this symbol 
does possess fractional properties, and later on we shall show how 
meanings may be attached to dy and dx, but for the present the 

symbol -~- is to be considered as a whole. 



DIFFERENTIATION 41 

Since the derivative of a function of 2; is in general also a func- 
tion of Xy the symbol /'(x) is also used to denote the derivative of 
f(x). Hence, if 

we may write -^ =zf*(x)y 

ax 

which is read the derivative of y with respect to x equals f prime 
of X. The symbol 

dx 

when considered by itself is called the differentiating operator^ and 
indicates that any function written after it is to be differentiated 
with respect to x. Thus 

-^ or -7- y indicates the derivative of y with respect to x ; 
dx dx 

-r-fi^) indicates the derivative oif{x) with respect to x; 
ax 

-- (2 a? -f 5) indicates the derivative of 2 a? 4- 6 with respect to x. 
dx 

The symbol D^ is used by some writers instead of — • If then 

ax 

y =/(^). 

we may write the identities 

43. Differentiable functions. From the Theory of Limits it is 
clear that if the derivative of a function exists for a certain value 
of the independent variable, the function itself must be continuous 
for that value of the variable. 

The converse, however, is not always true; functions having 
l)een discovered that are continuous and yet possess no derivative. 
But such functions do not occur often in applied mathematics, and 
in this hook only differentiable functions are considered, that is, func- 
tions that possess a derivative for all values of the independent 
variable save at most for isolated values. 



42 DIFFERENTIAL CALCULUS 

44. General rule for differentiation. From the definition of a 
derivative it is seen ttiat the process of differentiating a function 
y =^f{x) consists in taking the following distinct steps : 

General Rule for Differentiation 

First step. In the function replace x by x + Ax^ giving a new 
valiLe of the function^ if 4- Ay. 

Second step. Svhtract the given value of the function from the new 
value in order to find Ay (the increment of the function). 

Third step. Divide the remainder Ay (the increment of the func- 
tion) by Ax (fhe irusrement of the independent variable). 

Fourth step. Find the limit of this quotient^ when Ax (the incre- 
ment of the independent variable) approaches the limit zero. This is 
the derivative required. 

The student should become thoroughly familiar with this rule 
by applying the process to a large number of examples. Three 
such examples will now be worked out in detail. 

Ex. 1. Differentiate 3x< + 6. 

Solution. Applying the successive steps in the General Rule we get, after placing 

y = 3aj«+'5, 

First step. y + Ay = 3 (x > Ax)^ 4- 6 

= 3aja + 6x-Ax + 3 (Ax)8 + 6. 



Cy + Ay = 3x« + 6x-Ax + 
J y =3x« 



SecoTid step, jfy + Ay = 3x« + 6x • Ax + 3(Ax)* + 6 

+ 6 



Ay= 6x- Ax + S(Ax)*. 

Third step. — = 6x + 8Ax. 

^ Ax 

Fourth st^. Jl = 6x. . Ans, 

ax 

We may also write this 

-^(3x« + 6) = 6x. 
ox 

Ex. 2. Differentiate x> - 2 x + 7. 

Solution. Place y = ^ — 2x + 7. 

First step. y + Ay = (x + Ax)« - 2 (x + Ax) + 7 

= x» + 8x2 • Ax + 3x • (2to)2 + (Ax)« -2x-2.Ax + 7. 



DIFFERENTIATION 



48 



Second step. 



Third Oep, 



y + Ay = x« + 3x« . Ax + 3x . (Ax)« + (Ax)« - 2x 
y = x' - 2 X 



2- Ax + 7 
+ 7 



Ay = 8x2 . Ax + 3x • (Ax)« + (Ax)» 

^ = 8x2 + 3x • Ax + (Ax)2 - 2. 
Ax ^\ / 



-2 Ax. 



Fourth step. 


'' = 8x2-2. Ara. 
dx 




«'• ^t'^- 


2« + 7) = 3x2-2. 




Ex. 3. Differentiate — • 

xJ» 




Solution. Place 


c 

'^ X2 




First «tep. 


c 

t/ ^- Az/ ~~ ■ ' ' • 




• ' ' (x + Ax)2 




Seomd step. 


Ay- "" ^-" 


c-Ax(2x + Ax) 


"' (X + AX)2 X2 


X2 (X + AX)2 


Third step. 


Ay_ ^ 2x + Ax 
Ax~ ^ x2(x + Ax)2 




Fourth step. 

• 


dy_ ^ 2x_ 2c 
dx~ ^ xa(x)2~ x» 


Arys. 


Or, 


f_r,)=-J''. 





45. Applications of the derivative to Geometry. We shall now 
consider a theorem which is fundamental in all applications of the 
Differential Calculus to Geometry. Let 

y =/(^) 
be the equation of a curve AB. Consider a fixed point P whose 
coordinates are (a?, y). Let x take on an increment ^{=MN)\ 
then y takes on an increment Ay(=JBQ), the coordinates of Q 
being (a; 4- Aa^ y + Ay). ^ 

From the figure, MP = y =f(x) y 

and NQ^y-^Ay =-f{x 4- Ax); 

therefore EQ = Ay =/(a; 4- Ar) -/(a:). 

Draw a secant line through P and Q and 
a tangent line to the curve at P. Then 

^ ^ FE Ax Ax 

= slope of secant line FQS. 




N X 



I 



44 DIFFERENTIAL CALCULUS 

If we now let Aa: approach the limit zero, the point Q will move 
along the curve and approach nearer and nearer to P, the secant 
will turn about P and approach the tangent as a limiting position, 
and we may write 

^ limit Ay 
Aa; = Aa; 

__ limit f(x -h Ax) —f(x) 
"■A2:=0 Ax '^^' 

tan T = ^ from (F), p. 40 

= slope of tangrent line PT, Hence 

Theorem. The value of the derivative at any point of a curve is 
equal to the slope of the line drawn tangent to the curve at that point. 

It was this tangent problem that led Leibnitz* to the discovery 
of the Differential Calculus. 

Ex. 1. Find the slopes of the tangenU to the parabola y = sc^ at the vertex, and 
at the point where x = |. 

Sohdion, Differentiating by General Rtde, p. 42, we get 

dy 
(A) -^ = 2x = slope of tangent line at any point on curve. 

dx 

To find slope of tangent at vertex, substitute x = in (^), 

giving 3^ = ^• 

dx 

Therefore the tangent at vertex has the slope zero, that is, 
it is parallel to the axis of x and in this case coincides with it. 

To find slope of tangent at the. point P, where x = |, substi- 
tute in (^), giving 

dy _. 

dx'^' 

that is, the tangent at the point P makes an angle of 45^ with the axis of x, 

* Gottfreid Wilhelm Leibnitz (1G4&-1716) was a native of Leipzig. His remarkable abilities 
were shown by original investigations in several branches of learning. He was first to publish 
his discoveries in Calculus in a short essay appearing in the periodical Acta Eruditorum at 
Leipzig In 1684. It is known, however, that manuscripts on Fluxions written by Newton were 
already in existence, and from these some claim licibnitz got the new ideas. The decision of 
modern times seems to be that both Newton and lieibnitz invented the Calculus independently 
of each other. The notation used to-day was introduced by Leibnitz. 




DIFFERENTIATION 45 



Use the General Rule, p. 42, in differentiating the following examples. 

1. y-SxK 

2. y = x2-3x. 

3. y = 0x2 + ftx + c. 

4. y = x*. 
6. r = o^. 

6. p = 2g«. 

7. « =<« -2« + 3. 
a y = -. 

X 

9. « =-• 

10. Find the slope of the tangent to the curve y = 2 x* ~ 6 x + 5, (a) at the p^iiit 
where x = 1 ; (b) at the point where x = 0. Ans. (a) ; (b) — 6. 

11, (a) Find the slopes of the tangents to the two curves y = 3x^— 1 and 
y = 2 x^ + 3 at their points of intersection, (b) At what angle do they intersect ? 

Ana. (a) ±12, ±8; (b) arc tan ^. 



An,. f = 
dx 


6x. 




dy 
dx 


2x- 


3. 


dy 
dx 


2ax 


+ 6 


dy 

dx 


3x2. 




dr 
do 


2 0$. 




dp_ 
dq 


4q, 




da 
dt 


2t- 


2. 


dy 
dx 


1 

X2 


• 


ds 


4 




dt 


ti' 





CHAPTER VI 
RULBS FOR DIFFERENTIATING STANDARD ELEMENTARY FORMS 

46. Importance of General Rule. The Oeneral Rule for differ- 
entiation g^ven in the last chapter, p. 42, is fundamental, being 
found directly from the definition of a derivative, and it is very 
important that the student should be thoroughly familiar with it. 
However, the process of applying the rule to examples in general 
has been found too tedious or difficult ; consequently special rules 
have been derived from the 0-eneral Rule for differentiating cer- 
tain standard forms of frequent occurrence in order to facilitate 
the work. 

It has been found convenient to express these special rules by 
means of formulas, a Jist of which follows. The student should 
not only memorize each formula when deduced, but should be able 
to state the corresponding rule in words. 

In these formulas, u, t;, and w denote variable quantities which 
are functions of x, and are differentiable. 

FORMULAS FOR DIFFERENTIATION 

I ^ = 0. 

dx 

n ^ = 1. 

dx 

m —t A- — \ — — 4. — ^^!!£. 

dx ^ dx dx dx* 

--, d , . dv 

V —( \— —4- ^ 

dx "" dx dx* 

VI -j-iViVf'Vn) = (VtVf"Vn)'-=-^ + (VlV» • • • t?») -^ + ••• 

0nX/ 0iX €miX 

I t V dVn 

+ (t?iVi-..Vn-x)-=— • 

uX 

46 



RULES FOR DIFFERENTIATING 47 



vn 



YHa 



Yin 



Yma 



Yin 6 



n 



Ha 



la 



u 



m 



im 



nv 



IV 



IVI 



lYn 



^""l 


dx 


dx^ ' 


= nx^'K 


d /u\ 
dicKv) 


du dv 
V — — u — 

dx dx 




du 


d /u\ 
dx\c) 


dx 

C 




dv 


d(CX 

dx\vJ 


dx 




dv 


dx<*^^«^) 


dx 

= iogr«6 — • 

v 




dv 


^(loi,.) 


dx 

V 


dx^ ' 


«.i dv 


d . «v 


dx 


d . ^ 


= vu^'^— — I- logrti 
ax 


«te<^*"^> 


dv 

= C08V— — • 

dx 


—— (cosr) 
dx 


dv 

€207 


I<*»»«') 


« dv 

€207 


:^(cotv) 
dx 


• dv 

€207 


^(sect,) 


. dv 
= secvtanv—— • 

dx 


-— (cscv) 
ax 


^ dv 
= — cscvcott?-— • 

dx 



dv 
dx 



48 DIFFERENTIAL CALCULUS 



IVin -=— (vers v) = sin t;---* 

dx dx 

dv 

WW d . . . dx 

m -^— (arc sin t;) = 



dx Vl — V* 

dv 

__ d , V dx 

II -7-(arc cost;) = , • 

€iX "^1 — f;i 

in — (arctant;) = T -• 

^ dx^ ' l + v« 

dv 

Iin T- (arc cot V) = — T-^ — ;• 

dv 

d dx 

Um 3=^ (arc sect?) = 



dx V -y/v* 1 

dv 

d dx 

mV -T-(arcc8cv) = — 



dx V Vv* — 1 

dv 

^^„ d . V dx 

IIV -T- (arc vers v) = 



€to V2t; — 1;» 

jjYI ^ = ^. |!i, y i>eingr a function of v. 

dx dv dx 

IIYn w ~ w"' y beingr a function of x. 

uX €iX 

dy 

47. Differentiation of a constant. A function that is known to 
have the same value for every value of the independent variable 
is constant, and we may denote it by 

As X takes on an increment Az, the function does not change 
in value, that is. Ay = 0, and 

A:r 



RULES FOB DIFFERENTIATING 49 



But j^"'\(^)=^ = 0, 



I • — -o 

due 

The derivative of a constant is zero, 

9 

48. Differentiation of a variable with respect to itself. 
Let y = ^* 

Following O-eneral Rule^ p. 42, we have 
Firet step. y -f- Ay = a; -f- Aa;. 

Second step. Ay = Ax, 

Third step. t^=1- 

Ax 

FouHh step. -~ = 1- 

dz ' 

n .-. ^ = 1. 

dx 

The derivative of a variable with respect to itself ^^ unity. 
48. Differentiation of a sum. 

Let y = w 4- 1; — w. 

By General Rvlej 

First step, y4-Ay = u4-Au4-t;4-At; — tr — Am^, 

Second step. Ay = Au + Av — Aw, 

mr. •» ^ Av Aw . Av Aw 
Third step, t^ = t- + -i ir" 

TT ^T ^ dy du , dv dw 
Fourth step, ^ = — - + - --. 

ax dx dx dx 

[Applying Th. I, p. 27.] 

in • —( -L — ._dudv dw ^ 

dx, '^ dac dx dx* 

Similarly for the algebraic sum of any finite number of functions. 

The derivative of the algebraic sum of a finite number of functions 
is equal to the same algebraic sum of their derivatives. 



50 DIFFERENTIAL CALCULUS 

50. Differentiation of the product of a constant and a variable. 

Let y=cv. 

By General Rule^ 

First step. y -f- Ay = <?(v -f- Av) = «; -f- cAv. 

Second step. Ay = c • At;. 

Third step. --^ = {? -— . 

Ax Ax 

Fourth step. -^ = <• --. 

dx dx 

[Applying Th. II, p. 27.] 

_-- d , . dv 

IT .-. ^(«') = «5^- 

The derivative of the product of a constant and a variable is equal 
to the product of the constant and the derivative of the variable. 

51. Differentiation of the product of two variables. 

Let y = uv. 

By General Rule^ 

First step. y -f- Ay = (m -f- Au) {v -f- Av) 

^uv -\-U'Av '\-V' Au 4- Au • At;. 
Second step. Ay^u-Av-^-v- Au -f Au • Av. 

mi.' J * ^y Av ^ Au ^ . Av 

Third step. a=^^:— 4-v-- — hAu--—. 

Ax Ax Ax Ax 

rr ^1 ^ dy dv ^ du 

Fourth step. ^^u-^r-^-v- — 

dx dx dx 

C Applying Tb. 11, p. 27, since when ^x approachefl zero as a limit,"! 
Au also approaches zero as a limit, and limit r An — ^ j = 0. I 

„ d , . dv . du 

The derivative of the product of two variables is equal to the first 
variable times the derivative of the second^ plus the second variable 
times the derivative of the first. 

52. Differentiation of the product of any finite number of varia- 
bles. By dividing both sides of Y by uv^ it assumes the form 

d , . du dv 

— (UV) — — 
dx dx dx 

UV " u V 



I 



RULES FOR DIFFERENTIATING 61 

If then we have the product of n variables 



we may wnte 



VjV, • • • V, Vj v,v, • • • v^ 

_dx^ d^ dx^ ' * "^ 
dvj rfv, dv, dv^ 



dx dx dx dx 

Multiplying both sides by v^v^ • • • v«i we get 

YI :r-- (ViVi . . . V») = (ViV» . • . Vn) -^ + (VlV» • • • Vn) -^ + ••• 

dx dx dx 

ax 

The derivative of the product of a finite number of variables is equal 
to the sum of all the produMs that can be formed by multiplying the 
derivative of each variable by all the other variables. 

53. Differentiation of a variable with a constant exponent. If the 
n factors in YI ai*e each equal to t;, we get 

— (if) — 
dx dx 

— ,— . = n — • 
if V 

dx dx 

When v=sx this becomes 

dx 

We have so far proven VII only for the case when n is a positive 
integer. In § 59, however, it will be shown that this formula holds 
true for any value of n and we shall make use of this general result 
now. 

The derivative of a variable with a constant exponent is equal to 
the product of the exponent^ the variable with the exponent diminished 
by unity ^ and the derivative of the variable. 



62 DIFFERENTIAL CALCULUS 

54. Differentiation of a quotient. 
Let y = - > V ^ 0. 

V 

By General Rvle^ 

u4- Au 



First step. y + Ay = 



Second step. Ay = 



V -f Av 

u 4- At^ t^ V'Au — u-£iv 
v-f-Av v v{v-^Av) 



Third step. 



Fourth step. 



Au Av 
V tt — 

Ay ^ Aa; 
Ar t;(t;4-Av) 

du dv 

V u 

dy dx dx 

dx 1/^ 

[Applying Theorenu II and III, p. 27.] 

du dv 

V u — 

d /u\ doc doc 



Tin ... ^(»)=-^_ 

dx \v/ V 

The derivative of a fraction is equal to the denominator times 
the derivative of the numerator^ minus the numerator times the 
derivative of the denominator^ all divided by the square of the 
denominator. 

When the denominator is constant, set t; = c in Ym, giving 

du 

dx\e/ c 

We may also get YlUa from IV as follows: 

du 
d /u\ _ldu _dx 
dx\c J c dx € 

The derivative of the quotient of a variable by a constant is equal 
to the derivative of the variable divided by the constant. 



RULES FOR DIFFERENTIATING 63 

When the numerator is constant, set i^ = c in YIU, giving 

dv 

dx\v/ V* 

[sine ^-1-0.] 

The derivative of the quotient of a constant hy a variable is equal 
to minus the product of the constant and the derivative of the vari- 
able^ divided hy the square of the variable. 

All explicit algebraic functions of one independent variable may 
be differentiated by following the rules we have deduced so far. 



EXAMPLES 

Differentiate the following. 

1. y = x». 

Solution, ^ = -^(x^ = 3aj«. Am. by VII a 

dx dx 

ln-3.] 

2. y = ax*-6aJ». 

Sdidion. ^ = ^((ix*-6x«) = :^(aic*)-:^(te«) by III 

dz dz dx dx 

dx dx 

= 4ax»-26x. Ans, by Vila 

3. y = x* + 6. 

Solution, T^ = ~(«*) + :^(5) by III 

dx dx dx 

= ixK Ana, byVnaandl 

Solution. ^ = ^(3xV)-l(7ari) + -^(8x») by III 

dx dx dx dx 

=:J^x* + }x"*-f V»"*- -^'»*- by rv and Vila 

6. y = (a!^-3)». 

Solution, ^ = 5(xa-3)<^(xa-3) by VII 

dx dx 

[i? — X* -3 and n — 6.] 
= 5 (x2 - 3)* . 2x = 10x(x« - 3)<. Ana, 
We might have expanded this function by the Binomial Theorem and then 
applied III, etc., but the above process is to be preferred. 



64 DIFFERENTIAL CALCULUS 



6. y = Vd2-aJ«. 

Solutwn, ^ = :^(a^-x2)* = J(a«-x«r*^(a2-x«) by VII 

dz ax ax 

[©«a^-«« and n«-J.] 



7. y=(8x« + 2)VTT6i^. 

5ofu«ion. ^ = (3xJ» + 2)-^(l + 6a^)* + (l + 5x2)i;^(3xa4.2) by V 

dx ax ax 

IwSx* + 2 and o » (1 + 6j:>)i] 

= (3x« + 2) i (1 + Sx^)-* ^ (1 + 6x2) + (l + 6x«)*6x by Vn, etc. 

ax 

= (3x» + 2) (1 + 5x«r*6x + 6x(l + 5x2)* 

6x(8x» + 2),..,,r-i-r:;_46x. + i6x ^^ 

Vl + 5xa Vl + 6x« 

« a« + x« 

a y = 



Vaa-x^ 

(a« - x«)* -^ (aa + x«) - (a» + x«) ^ (a^ - x») * 

Solution. ^ = ^2 __ ^ byvm 

dx a2 - x2 

_ 2x(o« - X2) + x(a« + x2) 

(a* - x2)i 
(If oltiplying both numerator and denominator by (o^-x*)".] 



(a* - xa)i 



Arts, 



9. y = 5x* + 3x«-6. ^ = 20x« + 6x. 

dx 

10. y = 3cxa-8dx + 6e. --^ = 6cx-8(l. 

dx 

11. y = x«+». ^ = (a + 6)x«+»-i. 

dx 

dv 

12. y = x» + nx + n. -^ = nx*-* + n. 

dx 

13. f(x) = jx* - Jx^ + 6. /'(x) = 2x« - 3x. 

14. /(x) = (a + 6)xa + ex + d. /'(x) = 2 (a + 6)x + c 

d 

16. — (a + &x + cx«) = 6 + 2cx. 

dx 

16. — (6ir-3y + 6) = 5my«-i-3. 
dy 



RULES FOR DIFFERENTIATING 



55 



17. B = 



«0|>0 



1& V=:0o4-A 



19. s^so + oot + iyt*- 



20. l = l+W + c^. 



21. a = 2«« + 3t + 5. 



22. arrotP-W^ + c 



23. r = afi. 



24. r = c^ + <W» + c^. 



25. y = 6«' + 4x' + 2a'. 

26. y = V3« + Vx + -. 



27. y = 



a + te + cx^ 



28. y = ^ — r-^« 

X* 

x* - X - X* + a 



29. y = 



i 



30. y = (2x» + x«-5)«. 

31. /(x) = (a + ea?)*. 

32. /(x) = (l + 4x8)(l + 2x«). 

33. /(x) = (a + X) Va-x. 

34. /(x) = (a + x)'"(6 + x)-. 



35. y = — 

X" 



36. y = x(a9 + aJ2)Va«-x^. 





voPo 


do 


/. 


d8 


:»0+A 


dl 
d9 


:6 + 2ctf. 


dt 


:4t + 3. 


ds 
dt 


:3a^~2M. 


dr 
dB 


= 2a9. 



— = 3c^ + 2d« + e. 
dd 

^ = 21x« + 10x» + 3x*. 
dx 



•' + ' 



dy _ 

dy a 

-^ = c — -. 
dx xa 

^ = Jx>-6x« + 2x-* + ix-i 
dx ' 

dy _2x' + x + 2x*-8a 
dx" 



2x* 



dy 



^ = 6x(3x + 1) (2x« + aJ« - 5)« 
dx 

r(x) = ^(a + &x«)*. 



/'(x) = 4x(l + 8x + 10x«). 

a — 8x 
/'(x) = 



2 Va-x 
/'(«) = (a + x)".(6 + x).[^ + ^] 

dy 



dx 



n 

x« + i 



dy a* + a«x»-4x* 
*>! Va2 - x^ 



66 DIFFERENTIAL CALCULUS 

2«* dy 8Wx»-4z» 



37. y = 



38. y = 



39. a = 



6« - »a (te (62 - x«)« 

a — x dy _ 2a 

a + z' dx" (a + «)«" 

^ ds 3<« + <« 



(1 + «)« ctt (1 + 0« 

41. f(0)=—J=- /'W = 



Va-W« (a - W«)* 



44. «(x) = ^^^. *'(«)= '^"^ 



xVl + aJ» x»(l + ««)« 

dx L(o + x)"«(6 + x)"J (a + x)'»+i(6 + x)"+i' 



J* r ^g + g + ^g - g "1 _ __ g^ + aVg?"^^ 
^ Vo + x - Va-x xa Va« - x« 



JTiiif. Rationalize the denominator tint 

47. y=V2^. ^=?. 

dx y 

.Q 6^/-x 5 dy Wx 

a dx a?y 

49. . = («.-«.).. g = -^. 

d0 2 V0 

-, «<' + t7<' dtt v*'-* . v^-^ 

61. 14 = ^ ^. — = 1 

cd dv d (^ 



62. p=<?±l^. dp^ jq- 2) V^TT 

V^-l dg (g-1)* 



53. V = r — 5=7 

Ll + Vl - xa J 



dy ____ny__ 
*c ~ X Vl -. aja 



RULES FOR DIFFERENTIATING 67 

64. Given (a + «)* = a* + 6a*x + 10a«x«+ 10a^« + 6aa5* + «5 j find (o + x)* 
hy differeutiation. 

66. ABSummg that 

1 — x»+i 

= 1 +« + »«...+ x^, 

1 — « 

deduce by differentiation the sum of the aeriea 

l + 2x + 8«' + .-. + nx»-i, 

n being any positive Integer. 

nx»+*-(n + l)x» + l , 

^rri)5 • '^'^' 

55. Differentiation of a function of a function. It sometimes hap- 
pens that y, instead of being defined directly as a function of a;, 
is given as a function of another variable v which is defined as a 
function of x. In that case y is a function of x through v and 
is called a function of a function. 

For example, if y = r r, 

l — ir 

and v = 1 — a?, 

then y is a function of a function. By eliminating v we may 
express y directly as a function of x^ but in general this is not the 

best plan when we wish to find -^. 

ax 

If y =/(f ), 

and v=: <f} (x)y 

then y is a function of x through v. Let x take on an increment 

Ax, giving 

V -{-Av=^<f>{x + Ax)j defining Av, 

and y 4- Ay =f{v -f Av), defining Ay. 

By multiplying both numerator and denominator of — ^ by Av 

we get 

Ay Ay At' 

Ax Av Ax 

Let Ax approach the limit zero, then Av also approaches the 
limit zero, and we have,* applying Th. II, p. 27, 

dy^dydv^ 
' dx dv dx 

• Aflsumlng that Ao^O for Ax safficiently small bat not zero. 



58 DIFFERENTIAL CALCULUS 

This may also be written 

(B) ^= At') •*'(«?). 

If y = /(v) and v = ^(ic), the derivative of y with respect to x 
equals the 'product of the derivative of y with respect to v and the 
derivative of v with respect to xJ* 

56. Differeiitlation of Inverse functions. Let 

If the inverse function exists, denote it by 
{B) y ^f{x). 

Differentiating {B) with respect to y gives 

[AMmnlng ^(y) and/(a?) to be dlfferentlable.] 

-i ___dy dx 
' "^ dx dy 

dx 
If then —- = 0' (y) is different from zero, we get 

dy 

dy 

or, 

1 



(i» r(x) = 



The derivative of the inverse function is equal to the reciprocal of 
the derivative of the direct function, 
57. Differentiation of a*logarlthm. 

Let y = log^v.f 

Differentiating by the General Rvle^ p. 42, considering v as the 
independent variable, we have 

First step, y + ^y=: log„ (v + Av). 

* It is understood that y and r have fixed initial ralues corresponding to some fixed initial 
▼alue of X. 

t The student most not forget that this function is defined only for poeitiye values of the 
base a and the Tarlable v. 



RULES FOR DIFFERENTIATING 69 

Second step. Ay = log„(t; 4- Av) — log^v 



[By 8, p. 2.] 



r Dividing the logarithm by v and at the same time multiplying the exponent of thel 
[parentheslB by v chjuiges the form of the expression but not its value (see 9, p. 2). J 

du 1 

Fourth step. -ii = - log„«.* 

Am 

'when A9 approaches the limit sero, — also approaches the limit sero. 
Therefore ^^ v "*" — / ^^ ** *» '^^ Theorem II, p. 33, plaoing a s 

Hence 

Since v is a function of x and it is required to differentiate log^v 
with respect to a:, we must use formula (-4), § 56, for differentiat- 
ing B, function of a function^ namely, 

dy _ dy dv 
dx dv dx 

Substituting value of -^ from (A)^ we get 






dy 
dx 


= log.c 


1 (2t; 
t; dx 

dv 


n 


.*. 3-(l0gaV) : 

dx 


= lOga I 


dx 

'' V ' 


When a 


= e this becomes 


• 




Ha 


^(log«) = 


" V * 





* Since logflV is a continuous function. 



60 DIFFERENTIAL CALCULUS 

The derivative of the logarithm of a variable is equal to the product 
of the modulus* of the system of logarithms and the derivative of the 
variable^ divided by the variable. 

58. Differentiation of the simple exponential function. 

Let y^cL"' a > 

Taking the logarithm of both sides to the base e^ we get 

log y = w log a, 

or, v = --e^ = - logy. 

log a log a 

Differentiate with respect to y by formula Ha, 

^ = JL 1. 
dy^loga y* 

and from (C), § 56, relating to inverse functions^ we get 

^=loga.y, or, 

(.1) ^ = loga.a''. 

Since t; is a function of x and it is required to differentiate a" 
with respect to 2:, we must use formula (A)^ § 55, for differenti- 
ating a function of a function^ namely, 

dy __dy dv 
dx dv dx 

Substituting the value of -^ from (-4), we get 

dy y „ dv 

-f- = log a •a'' -5-. 
dx dx 

w <f . V - dv 

I ••. -T-Ctt*) = loera •«»•-—-• 

doD dx 

When a = «, this becomes 

la —-(«») = c*'---. 

daC' dx 

•The logarithm of « to any base a (= loggC) is called the modulus of the system whose base 

is a. In Algebra It is shown that we may find the logarithm of a number N to any base a by 

means of the formula . j^ 

log«iV= logac . logeA'= 

lofea 

The modulus of the common or Briggs* system with base 10 is 

logioe«. 434294 •". 



BUL£S FOR DIFFERENTIATING 61 

The derivative of a constant with a variable exponent is eqtuU to 
the product of the natural logarithm of the constant^ the constant with 
the variable exponent, and the derivative of the exponent, 

59. Differentiation of the general exponential function. 

Let y = tt'.* 

Taking the logarithm of both sides to the base «, 

log.y=vlog.tt, 
or, y =«"'*«•. 

Differentiating by formula Xa, 




by¥ 



dx , dv ] 



n ... A (t^f) -. ^u^-i^ + log u . i«« ^. 

dx dx dx 

The derivative of a variable with a variable exponent is equal to 
the sum of the two results obtained by first differentiating by YII, 
regarding the exponent as constant; and again differentiating by I, 
regarding the base as constant. 

Let t; = n, any constant ; then XI reduces to 

But this is the form differentiated in § 68, therefore YII holds 
true for any value of n. 



Ex. 1. Differentiate 


y = log(a5» + a). 


Sohitioii. 


dy dx 

dx x^ + a 




[r=a:«+a.] 




2x 
= -r An». 




• 11 can here assume only positive ralues. 



bylXo 



62 DIFFERENTIAL CALCULUS 



Ex. 2. Differentiate y = log Vl - z\ 






Soltrfioj*. ^ = — 

d* (1 _ a;!i)J 




bylXa 


_i(l -*»)"*(- 2x) 
(1 - x«)» 




by VII 


■^ i^ j<4«jt 






aj»- 1 




Ex. 3. Differentiate y = a»«*. 






Solution. ^^ = log o . a»«^ f (8 x«) 

dx dx 




byX 


= 6xloga-a««". a-m. 






Ex. 4. Differentiate y = 6c«"+«*. 






Solution. ^*^ = 6^ (c^+«^) 

dx dx 




by IV 


= 66^ + «*^ (c«+x«) 
dx 




by Xa 


= 26xe«^+«". 2ln«. 






Ex. 5. Differentiate y^if. 






SohOUm. f ^ = (FTf^- » ^ (X) + x** log x ^ Je*) 

dx dx d«^ 


by XI 


= e^Tf^^ + x** logx { c*j 






= e»*^(- + logxj. ilna. 





60. Logarithmic differentlatioii. Instead of applying H and 
IX a at once in differentiating logarithmic functions, we may 
sometimes simplify the work by first making use of one of the 
formulas 7-10 on p. 2. Thus above Ex. 2 may be solved as 
follows. 



Ex. 1. Differentiate y = log Vl — x". 

Solutum. By using 10, p. 2, we may write this in a form free from radicals as 
follows. 

y = i log (1 - aJ»). 

Then -^ = by IX a 

dx 2 1-xa ^ 

1 -2x X . 

= ::'z :: = -z • -^n*. 

2 l-x2 xa-1 



BULES FOR JOIFFERENTIATING 63 



Ex. 2. Differentiate y 



1 /nn? 



Schition. Simplifying by means of 10 and 8, p. 2, 

V = iPog(l + x«) - log(l - ai«)]. 



2L l+x« l-«« J 



dy 

dx ^_ 

*" 2x 



by IX a, etc. 



Am. 



l+«» \-7fi l-«* 

In differentiating an exponential function, especially a variable 
with a variable exponent, the best plan is first to take the loga- 
rithm of the function and then differentiate. Thus Ex. 5, p. 62, 
is solved more elegantly as follows. 

Ex. 3. Differentiate y = x«*. 

Solution, Taking the logarithm of both sides, 

log y = e« logx. By 9, p. 2 

Now differentiate both sides with respect to x. 

dy 

— = e*— (logx) + logx-^ (6«) by IXo and V 

y dx dx 

= e« — h log X • e*, 
or, ^ = «'-y(- + lo««) 



dx 



= cx^(- + logx). An^ 



Ex. 4. Differentiate y = (4 x« - 7)« + '^'^^. 
Solviion, Taking logarithm, 

logy = (2 + Vxa-6)log(4x« - 7). 
Differentiating both sides with respect to x, 

1J? = (3 + V^36) 8^+log(4*«-7).-^ 
ydx 4x«-7 Vx«-6 

*? _-/.-., ^v^%^Sr7r8(2 + Vx2-6) . log(4x«- 
dx 



= x(4xa-7)-^^r«^^±^^ ^n,. 

L 4x^-7 VSTTg J 



In the case of a function consisting of a number of factors it is 
sometimes convenient to take the logarithm before differentiating. 
Thus, 

Ex. 6. Differentiate y = ^(^-^)(^-^) . 

\(x-3)(x-4) 

Solution. Taking logarithm, 

logy = i[log(x - 1) + log(x - 2) - log(x - 8) - log(x - 4)]. 



64 DIFFERENTIAL CALCULUS • 

Differentiating both sides with respect to x, 

1 dy _ 1 r 1 ^l 1_ _ 1 n j 

2ga-10x + ll 

(X - l)(x - 2) (X - 8)(x - 4)* 

dy 2xa-10x-U 
or, -=- = . An», 

*B (X - l)» (X - 2)* (X - 3)*(x - 4)* 



Differentiate the following. 
1. y = log(x + a). 



EXAMPLES 



dy 



2. 


y = log(ax + 6). 


3. 


y = log- 

1 — X 


4. 


, l+x« 


6. 


y = e^. 



dx x + a 
dy a 



6. y = e**+ 



7. y = log (x« -f X). 



8. y = log(x»-2x + 6). 



dx 


ox + & 


dy_ 
dx 


2 

l-x* 


dy 
dx 


4x 
1-x* 


dy_ 
dx 


ac«. 


dy_ 
dx 


4c<*+«. 


dy_ 


2x + l 



dx x^ + x 
dy 3x« - 2 



dx x* — 2x + 6 



9. y = log. (2 « + «•). 5f = log„e|±i^. 

dx 2x + x> 

10. y = xlogx. J? = iogx + l. 

dx 

11. /(x) = logx». /'(x) = ?. 

X 

12./(x) = log»x. /'(x) = ii?i!^. 

X 
Hint. log«ar«(logar)». Use tint Vn, v» logz, n= 3 ; and then IX a. 

13. /(x) = log?^l^. /'(x)= ^^ 



a - X a* - x^ 



14. /(x) = log (X + vT+x5). f (X) = -A 



Vl + x« 



BULES FOB DIFFEBENTIATING 



65 



16. y = c^. 

16. y = b^. 

17. y = 7**+«*. 

18. y = c«*--^. 

19. r = o«. 

20. r=za^: 

21. a = «*■+**. 

22. u = cu/^, 

23. p = e«>«««. 



dz 



= log a • o«*e*. 



^ = 2«log6.6»». 



dy 
dx 

dy 
dx 



= 21og7(x+l)7«^+««. 



= — 2xlogc«c«*-«^. 



d$ ^ 

dr __ g}^ ^ log a 
de~ $ 

^ = 2te6"+^. 
dt 

du ae^ 
d» 2 Vc 

^ = e9i«f«(l + log5). 



24. f-[e'(l-x«)] = c«(l-2« 
dz 



26. l/glz2)= 2^ . 



-«»). 



26. — (x«c«') = Jce<«(ax + 2). 
dz 



27. y = log 



e« 



1 + e« 



28. y = |(cS-c »). 



29. y = 






30. y = z^a/'. 



31. y = x*. 



32. yrra^. 



33. y = xJ<«*. 



dx~"H-c*' 
d2^ 4 



dz 


(ex + c-.,a 


dy 
dz 


:a*x»-Mn + xl 


dy^ 
dz 


:X*(l0gX + l). 




1 


dy 


x*(l — logx) 


dz 


x« 


dy 
dz 


:10gXa.X»<«*-*. 



66 DIFFERENTIAL CALCULUS 

84. /(y) = logy.«i'. /(y) = ei'(logy + i) 

1 



36. /(x) = log(log«). /(«) = 



aclogx 



• 37. J'(*) = log«(logx). y(^)^ 41og»(logx) 

xlogx 

88. 0(x) = log(log*«). 0'(«)= * 



xlogx 



89. ^(y) = log.JlI|. ^'(y) = _-L_. 

40. /(x) = log:^^^^l^. /'(x) = - ^ 



Vxa + 1 + X Vl +x« 

JTini. fint rationalise the denominator. 



41. y = x^««*. ^ = 0. 

dz 

42. y = e^. -?^ = €*•(!+ log x)x«. 

ax 

d© \ t> / 

«• «=(i)' |=(f)W«-log.-l). 

47. y = x«^. --^ = x«*+"-i(nlogx + l). 



dx 



-?? = x«"x«(^logx + log«x + - V 
dx \ x/ 



48. y = x*". 

49. y = a^^5=5. ^= xylogg 

^ (a« - x2)* 

50. y = c*(x»-nx«-i + n(n- l)x"-* ). -?^ = c«X». 

dx 

61 (X + 1)« tfy_ _ (x + l)(6x« + 14x + 6) 

(x + 2)»(x4-3)** dx (X -H 2)* (X + 3)* 

.ERn^ Take logarithm of both sides before differentiating in this and the following examples. 



RULES FOR DIFFERENTIATING 



6T 



ICO <* - !)• 
62. y = — ^2 ^ . 

(«- 2)*(x-3)« 
68. y = xVl-x(l+«). 

66. y = x»(a + 3x)«(a-2»)«. 



dy 
^ 



dy _ (X - l)»(7x« + 80x - 97) 
*» 12(x-2)*(x-8)V 

dy 2 + x — 5x* 
*» 2 Vl-x 
dy_ 1 + 3X^-2x* 
<to " (1 - x»)* 

= 6x* (a + 8x)«(o - 2x) (a« + 2ax - 12x«). 



66. y = 2g±i!. 
Vx~o 



dy (x — 2 a) Vx + o 
^ (X - a)* 



61. Differentiation of sin v. 

Let y = sin t?. 

By General Rule^ p. 42, considering v as the independent 
variable, we have 

First step. y + Ay = sin (v + ^v)* 

Ay = sin (t; 4- Av) — sin t^ 

^ / AvX . Av ♦ 
8= 2 cos/ v + — J . sin — . 

/. Av^ 



Second step. 



Third step. 



Av 
2 



Fourth step. 



Slnoa 



and 



and 



•Let 

I 




Adding, 
Therefore 


A-¥B^2v\-^v 

2 



-^ = cos V. 




^-o('^Vi^a4).p.3o. 




\ -sT / 








^"« + Av 




B^v 


Suhtneting, ^ 


i-B^Av 


\{A 


-*-T- 



Sabttitnting theie valnes of ^, J7, i(^ + ^, 4(^-1?) In terms of v and Av in the formula from 

Itlgonometry (42, p. 8), 

■in^-sln^-2coeiM + ^8ini(^-^, 



we get 



iln(» + Ar)-ilnp»2ooefr + -^8ln— -. 



68 DIFFERENTIAL CALCULUS 

Since v is a function of x and it is required to differentiate sin t; 
with respect to x, we must use formula (^), § 55, for differentiating 
^function of a function^ namely, 

dy _dy dv 
dx dv dx 

Substituting value -~ from Fourth step^ we get 

dy dv 

-f- =cos V-—- 
dx dx 

HI ••• — - (sin v) = cos V —— • 

iix ax 

The statement of the corresponding rules will now be. left to 
the student. 

62. Differentiation of cos v. 

Let y = cos v. 

By 29, p. 2, this may be written 

Differentiating by formula III, 



dy 

-2. — cos 

dx 



= cos( 

dv 

= — sm V ^- 

dx 



i-")(-S) 



Fsinoe COB (-- v\ *» Bin v, by 29, p. 2.1 



IIII ••• ^— (cos v) = — sin V— — 

ax ax 

63. Differentiation of tan v. 

Let y = tan v. 

By 27, p. 2, this may be written 

sin V 

y = 

cost; 



I 

RULES FOR DIFFERENTIATING 69 



Differentiating by formula vm, 



cos v ^- (sin v) — sin V ^- (cos v) 
dy dx^ ' dx^ 

dx cos^v 

^ dv , . i dv 

cos* V -r- + 8in' V -r- 

dx dx 



cos* V 



dv 



dx a dv 

= sec V — . 



cos* V dx 



d, . dv 

HT .•. ^ (tan V) = 8ec«t? — • 

64. Differentiation of cot v. 

Let y = cot V. 

By 27, p. 2, this may be written 

1 

y = T 

tan V 
Differentiating by formula YIII 6, 



dx tan* V 



IT ... ^(cot*) = -«««*»'rf^ 

65. Differentiation of sect;. 

Let y = sec v. 

By 26, p. 2, this may be written 

1 

y = 



, dv 
sec* V -r- , 
dx o at^ 
= — csc^ V— . 

tan*t; aaJ 

dv 



cos V 



To biFFEBENtlAL CALCULUli 

Differentiating by formula Yin 6, 

dx cos^v 

dv 
sint;—- 

dx 



cos* v 



, dv 
= sec vtant;--- 

dx 

___ ' d , dv 

lYI .•• :T-(8ect^) = sect; tan t7---« 

66. Differentiation of esc v. 

Let y = CSC v. 

By 26, p. 2, this may be written 

1 

y = . 

sin V 
Differentiating by formula YUI 6, 

d , , . 

(2a; sin* v 

dv 

COS v^- 

_ (fa 

sin*v 

, dv 

= — CSC vcot V--- 

(fa 

IVn ••• -r-(C8Ct;)= — CSCVCOtV—^. 

ax die 

67. Differentiation of vers v. 

Let y = vers v. 

By Trigonometry this may be written 

y = 1 — cos V. 
Differentiating, 

dy . dv 

-:p- = sin v — - • 
cea: dx 

ITm .% :^ (vers r) = sin v ^. 

dx dx 



RULES FOR DIFFERENTIATING 71 

In the derivation of our formulas so far it has been necessary 
to apply the General Rule^ p. 42 (i.e. the four steps), only for the 
following : 

TTT ^ / . . du dv dw A, , . 

Ill -r-(M + V — f£?) = -7--|--; — 3-- Algeoraic sum. 

ax dx dx ax 

V -r-(^v) = u-7--|- V-;-- Product. 

dx dx dx 

du dv 

VIII ± (1\ = !5I^. Quotient. 

dx\vj tr 

dv 

d dx 

IX — (log^ v) = log.e — • Logarithm. 

XII -;-(sin v) = cost;-7- Sine, 

aa; dx 

XXVI ^ ^ ^ rftj Function of a function. 

dx dv dx 

Not only do all the other formulas we have deduced depend on 
these, but all we shall deduce hereafter depend on them as well. 
Hence it follows that the derivation of the fundamental formulas 
for differentiation involves the calculation of only two limits of any 
difficulty, viz., 

S'-T-'=i by (14), p. 30 

and J^J (1 + ")* = «• By Th. II, p. 88 



Differentiate the following. 
1. y = sin ox^ 

^ = co8ax«~(ax^ ^r XII 

dz ax ^ ' 

= 4axco«i?a^V 



72 DIFFERENTIAL CALCULUS 

2. y = tan Vl — x. 

^ = 8ec2Vrri-^(l-x)i by XIV 

ox ax 



sec^ Vl - X 



2Vl-« 

* 

3. y = cos^ 

This may also J)e written 
y = (cos x)'. 

^ = 3 (co8x)«-^ (coB«) by VII 

ax ax 

iv » COS X and na 3.] 
= 8 cos* X ( - sin x) by Xni 

= — Ssinxcofi^x. 

4. y = sin nx sin" x. 

— = sinnx— (8inx)» + »in»x-'(8innx) by V 

dx dx dx ' ^ 

[tt— Bin nor and r » Bin" or.] 

(J /f 

= sinnaj- n(sinx)"-» — (sinx) + sin"xcosnx— (nx) by Vn and XII 

ax dx 

= nsinnx-8ln»-'xco8x + n sin" x cos nx 

= n sin"-! x(8in tix cos x + cos nx sin x) 

= n sin»- 1 X sin (nH- 1) £• 

e dy 

5. y = secax. -^^asecoxtanox. 

dx 

6. y = tan(ax + 6). ~ = a sec* (ox + 6). 

dx 

7. y = sin«x. — = sin2x. 

dx 

S. y = co8»x». -^ = -6xcos*x«sina^. 

dx 

9. f(y) = sin 2 y cosy. /'(y) = 2 cos 2y cos y - sin 2 y sin y. 

10. F(x) = cot* 5 X. JP''(x) = - 10 cot 6x cosec* 6x. 

n. F(^) = tan^-^. F'(^ = tan9^. 

12. /(0) = 0sin0 + cos0. /'(0) = 0co8 0. 

13. /(Q = sin* e cos t. /' (Q = sin^ t (3 cos» e - sin* Q. 

14. r = a cos 2 ^. — = — 2 a sin 2 tf. 

d0 



RULES FOR DIFFERENTIATING 



73 



16. r = aVcoe2e. 



16. r = a (1 — C08 ^. 



17. r = a8iii«-. 

3 

18. —- flog cos x) = — tan X. 
dx 

19. — Oogtanx) = -r-;--. 
dx 8m2x 



dr aain2$ 
d^ Vco82tf 

— = a sin 6, 
d0 

dr . o^ ^ 

— = a sin* -COS-. 

d0 3 3 



^' 3-(logsln«x) = 2cotx. 
dx 



21. y = 



tanx — 1 
secx 



, /l + sin X 

22. y = log-^- ; 

\ 1 — sin X 

23. y = logtanQ + |). 

24. /(x) = 8in(x + a)co8(x — a). 
26. /(x) = 8in(logx). 



26. /(x) = tan(logx). 



27. « = cos-' 



28. r = 8inl. 



29. p = sin (coe q). 



30. y = c^n* 



31. y = o*«>"*. 



32. y = e«»*8inx. 



dy 

-=^ = sin X + COS X. 
(ix 

dy 1 



dx 


cosx 


dy 


1 


dx 


cosx 


/'(«) = 


cos2x. 


/'(3:) = 


cos (log x) 


X 


/'(x) = 


sec* (log x) 


X 


da 


. a 
asm-- 


dt 


«2 




1 


dr 


2C08-- 



d& ^ 

-^ = — sin 9 COB (cos q), 
dq 

— = C^n*COSX. 

dx 

J^ = na*"** sec* nx log a. 
dx 

_K = 00012 ^cos X — sin* x). 
dx 



33. y = e* log sin x. 



dy 
dx 



= c* (cot X + log sin x). 



74 



DIFFERENTIAL CALCULUS 



34. JL (x«c^naf) = a?»-' e^*(ii + x cos x). 
dx 



35. — (e««coBmx) = 6«*(aco8m» — mslnm*). 
dz 



36. /($) = 



1 + CO8 
1 — COB0 



2 sin 9 



^,^v _ c°»(a Bin »- cos ») 

38. /(«) = (« cot «)». 

39. r = |tanS0-tAn0 + tf. 

40. y = a^«. 

41. y = (sin x)*, 

42. y = (sinx)««*. 

43. y = x + logcosTx - ^V 



J v/ — 


(1 - cos 6)^ 




/'W = 


:€9^Bin^. 




/'W = 


28cots(cot« — a 


cosec* a). 




tan«tf. 




dy^ 
dx 


. ^/sinx . , 

X«*n*( 1- log 

\ X 


XCOBxY 



dy 

-^ = (sin x)* [log sin X + X cot x]. 

dx 

- - = (sin x)*^*(l + sec* x log sin x). 
dx 



dy 



2 



dx 1 + tanx 

44. From sin 2 x = 2 sin x cos x, deduce by differentiation 

cos 2 X = cos* X — sin* x. 

. n+1 , nx 
sin— -—xsin — 
2 2 

45. From sin x + sin 2 x + • • • + sin nx = :: ♦ deduce by differentiation 



cosx + 2cos2x+"* +n cos nx = 



sin- 



n+1 , X . 2n+l 

sin - sm x 

2 2 2 



-K 



sin 



n + 1 



)* 




sin*- 
2 



[n«- a poBitiTe integer.] 

68. Differentiation of arc sin v. 



Let 
then 



y = arc sin v ; * 
V = sin y. 



* It Bhould be remembered that this function is defined only for raluef of v 
between - 1 and + 1 IncluBive and tliat p (the function) ii many-Talued, tliere 
being infinitely many arcs whose sines all equal v. Thus, in the figure (the locus 
of y =s arc sin r), when r = OAf, y - AfPj, Af/*„ Af/>„ • • •, Af ^j, MQ^, ••-. In the 
above discussion, in order to make the function slngle-ralued, only yaluee of y 

between -' and ' indusiye (points on arc QOP) are considered; that ISj^the arc 

z ^ 
of smallest numerical value whose sine is v^ 



kULES ^OR btFFfe&ENtlATti^Gt 



Yfi 



Differentiating with respect to y by in, 

dv 

— = cosy; 

dy 

dy_ 1 

dv cos y 
But since v is a function of x^ this may be substituted in 



therefore 



By (C), p. 68 



giving 



dy __ dy dv 
dx dv dx 

dy 1 dv 
da; cosy dx 

1 rfr 



(-4), p. 57 



-yflZr^dx 

[000 y-" VI -Bin* y - v^l-r»; the positlye sign of the radical being taken*1 
since cosy is positive for all values of y between -^ and ^ inclusive. I 



HI 






d 



dv 
doc 



^ (arc sin v) = - ^ 

doc Vl — v« 



69. Differentiation of arc cos v. 

Let y = arccost;;* 

then t; = cos y. 

Differentiating with respect to y by nn, 

dv 
dy 

di^ 1 



= - sin y ; 



therefore 



dv 



siny 



By (C), p. 68 



- Y 



But since t; is a function of 2:, this may be substi- 
tuted in the formula 

dy __ dy dv 
dx dv dx* 



{A), p. 57 



* This function is defined only for values of r between -1 and + 1 inclusive, 
and is many-valued. Ii^the figure (the locus of y»aroco8v), when vOM, 

In order to make the function single-valued, only values of y between 
and ir inclusive are considered ; that is, the smallest positive arc whose cosine 
is V. Hence we oonfine ourselves to arc QP of the graph. 




76 



DIFFERENTIAL CALCULUS 



giving 



dy 1 dv 

dx siny dx 



dv 



Vl-v*^ 



XX 



[Blny-%^l-ooi*y=v^l-i>*, the pluB sign of the radical being takenl 
Blnoe ainy ia poaitive for all yaluea of y between and v inoli]alYe.J 

dv 

d , doc 

••• -— (arc cos v) = • 

doc Vl — V* 



70. Differentiation of arc tan v. 
Let y = arctant>;* 

then V = tan y. 

Differentiating with respect to y by XIY, 

dv , 

— = sec"y; 
dy 



therefore 



dy^ 1 

dv sec* y 



By (C), p. 68 



But since v is a function of x^ this may be substituted in the 
formula 



giving 



dy dy dv 

dy 1 dv 
(2i; sec'y (ir 

1 dv 
\-\''^ dx 

[aec* y = 1 -I- tan* y « 1 + «•.] 



(^), P- 57 




XXI 



d_ 
dK 



(arctant;) = 



dv 
dx 



l+r« 



• This function ia defined for all yaluea of v and is many- 
valued, aa is clearly shown by its graph. In order to make it 

single-yalned, only yaluea of y between- \ and ^ are conaid- 

ered ; that ia, the arc of smalleat numerical yalue whose tangent 
lav (branch ^02))* 



RULES FOR DLFFERENTIATING 



77 



7L Differentiation of arc cot t;.* 
Following the method of last section, we get 

dv 

xin 



<i . .V doc 

•-r-(arc cot V) = — -r-- — -. 
dao 1 + 1>* 



72. Differentiation of arc sec v. 

Let y = arc sec v ; t 

then t; = sec y. 

Differentiating with respect to y by ITI, 

dv 

— = secytany; 



therefore 



dv sec y tan y 



By (C), p. 68 



But since v is a function of sc^ this may be substituted in the 

formula 

dy dy dv 



(A), p. 57 



* Thii fnnetion to defined for all yalues of v and 1b many-Talued, aa la leen from Its graph 
(Fig. a). In order to make ft slngle-yalned, only raluee of y between and ir are ooniidered ; 
that is, the smallest positive arc whose cotangent is v. Hence we confine ourselyes to branch AB, 





Flo. 6 



t This function is defined for all yalnes of v except those lying between -1 and + 1, and Is 
seen to be many-valued. To make the function slngle-ralued, y is taken as the arc of smallest 
numerical value whose secant Is v. This means that if v is positive we confine ourselves to 

points on arc AB (Fig. b), y taking on values between and | (0 may be Included) ; and if o is 

negative we confine ourselves to points on arc DC^ y taking on values between -v and ~- 
(-« may be included). 



78 



DIFFERENTIAL CALCULUS 



giving 



av 



xim 



dx sec y tan y dx 
1 dv 

t; Vv^— 1 rf^ 

riecy-r,Midtaiiy- v^8eo«y-l=^r»-l, the pins gig;]! of the"! 
I radical being taken Binoe tan p 1b poBltiye for all yalnea of y I 

[between and ~ and between - v and - ^ , including and - v. I 

dv 
d , . dx 

die t, Vv« — 1 



• • 



73. Differentiation of arc esc v.* 

Following method of last section, 



my 



d 



dv 
dx 



— - (arc CSC v) = — . 

dx t?Vv* — 1 



74. Differentiation of arc vers v* 

Let y = arc vers t;;t . 

then V = vers y. 

• Thla function 1b defined for all yaluee of v except those lying between -1 and +1, and ia 
seen to be many-valued. To make the function single-yalued, y is taken as the arc of Bmalleat 
numerical ralue whose secant is v. This meana that if v is posltire we confine ourselres to 
points on arc AB (Fig. a), y taking on values beween and ^/f may be included\ ; and if v is 

2\2 / , 

negative we confine ounelves to points on arc CD, y taking on values between -v and -- 

( - ^ may be indudedV 



2 





Fio. a 



FIO. ft 



t Defined only for values of v between and 2 induBlve, and is many-valued. To make the 
function continuous, y is taken as the smallest positive arc whose versed sine is v ; that is, y lies 
between and v inclusive. Hence we confine ourselves to arc OP of the graph (Fig. 6). 



RVLES FOR DIFFERENTIATING 79 

Differentiating with respect to y by XYm, 

dv 

therefore f = ^' By (C), p. 58 

But since v is a function of Xj this may be substituted in the 
formula 

dy 1 dv 

giymff -^ = 

dx Biny dx 

1 dv 



^2v — v^^^ 



tfbi y * Vl-coB*tf " ^l-(l-veray)« = v^2 1? - r», the plus sign of the radical beingl 
taken Blnoe Biny la positive for all values of y between and v Inclusive. J 

dv 

d , ^ doc 

HV .•. -— - (arc vers v) = 



doc V2t; — r* 



Differentiate the following. 
1. y = arctanax>. 

SolvJUm. cgf^ «* by XXI 





(iy_ 


i^'^ 




dx 


1 + (ox*)* 




= 


2ax 

l + a^x* 


-4»«) 


• 




dy__ 


^(8x 4«.) 


dx 


vr: 


(3x- 4x8)3 




[» 


-3a;-4a<.] 


, 




3-12x« 



2. v = aicain(3x — 4x'). 



A>iii£i(m. ?= byxix 



3 



Vl - 0x2 + 24x*- 16x0 Vl-x^ 



80 DIFFERENTIAL CALCULUS 

x^ + l 



3. y = arc sec 



dx\za-l/ 



Solution. ^ = ^ ;'" "^ by XXm 

xa_i\\xa-l/ 

(x«--l)2z-(a;a + l)2g 

(xa - 1)« 2 



4. y = arc sin - 



x^+j. 2x x« + 1 

x2-l*xa-l 

dy 1 



6. y = arc cot (x* — 5). 



a dz Va« - x* 

dy — 2x 



6. y = arc tan 



dz 1 + (x* - 6)a 
2x dy 2 



7. y = arc cosec 



1 - x^ dx 1+ x« 

1 dy 2 



8. y = arcyer82x'. 



2x«-l dx Vl-x? 

dy 2 



9. y = arc tan Vl — x. 



dx Vl - x« 
dy 1 



<^ 2 Vl-x(2-x) 



lA 8 dy 2 

10. y = arc cosec — " 



11. y = arc vers 



2x dx V9-4x* 

2x3 dy 2 



1 + x« dx 1 + xa 



io * 35 dy a 

12. y = arc tan - * " 



13. y = arc sin 



a dx a^ + z^ 

. x + 1 dy_ 1 



V2 d* Vl -2x-x« 

14. /(x) = X Va2 - x3 + a^arcsin -. /'(x) = 2 Va« -x^. 



a 



a — x\l 



16. /(x) = Va* - x2 + aarcsin -. /(x) = ("^^^^^ 

a \a + x/ 



16. X = r arc vers?- — V2 ry — y*. 



dx 



17. tf = arc8in(8r-l). 



»• dy V2ry-y* 

d^ 3 



dr V6r-9r* 



18. = arc tan 



19. 8 = arc sec 



RULES FOR DIFFERENTIATING 81 

r + g d»_ 1 

1 -or' dr ""l + f«* 

1 da 1 



Vl -t* <tt Vi-t« 



d X 

20. -—(xarcsinx) = arc8iIlx + 



<*3t Vl-x* 

^ d o tan tf 

21. — (tan $ arc tan 0) = sec* ^ arc tan -\ — - 

da^ ' 1+^ 

d 1 

22. —[log (arc cos i)] = 



dt arc cos t Vi — f-* 

23. /(y) = arc cos (logy). f(y) = - 



y Vl - (log y)a 



24. /(d) = arc sin Vsind. /(«) = ^ vTTcosecS. 






+ CO8 
26. p = C^*"9. 



27. u = arc tan 



28. 8 = arc cos 



dq l + g« 
e" — e-» du 2 



2 du C + r-T 

c« - e-* d« 2 



e' + c-' dt c< + e-' 



.^ dy . /arcsmx . logx \ 

dx \ X Vi - x*'^ 

30. y = c**arctanx. -J^^c^f:; - + x* arc tan x(H- logx) |. 

dx Ll + x* J 



31. y = arc sin (sin x). T" = ^• 



dx 



^4 sin X 
32. y = arc tan 



3 + 6 cos X 



a , jx — a 

33. y = arccot- + log\/ 

\x + a 



34. 



X lfX + 

l+x\l 1 



36. y = Vl - x' arc sin x — x. 

x«»~l 



36. y = arc cos 



dy 


4 


dx 


5 + 3 cos X 


dy 


2ax* 


dz 


X* — a* 


dy _ 


X2 


dz 


1-x* 


dy _ 


X arc sin x 


dx 


Vl-x2 


dy 


2nx"-i 



aj2" + 1 dx x2" + 1 



82 DIFFERENTIAL CALCULUS 

Formulas {A), p. 57, for differentiating 9^ function of a function^ and (C), p. 58, 
for differentiating inverse fanctionSj have been added to the list of formulas at the 
beginning of this chapter as XXVI and XXVII respectively. 

In the next eight examples, first find -^ and —- by differentiation and then 

uv dx 
substitute the results in 

*? = ^.* by XXVI 

dv • dz dv dx 

to find -^ . 
dx 

In general our results should be expressed explicitly in terms of the independent 

variable ; that is, — in terms of x, — in terms of y, --^ in terms of 0, etc. 

dx dy d$ 

37. y = 2t>«-4, c = 3x« + l. 

— = 4«; ~ = 6x; substituting in XXVI, 
dv dx 

^ = 4c.6x = 24x(8x« + l). 
dx 

38. y = tan 2 v, v = arc tan (2 x — 1). 

— = 2seca2t; — = ^ ; substituting in XXVL 

dv 'dx 2x2-2x + l ® ^ 

dy_ 2sec«2p tan'2i?-fl _ 2xg-2x-fl 

(ix""2x*-2x + l ~ 2x2-2x + l~ 2(x-x2)2 

Since va arc tan (227-1), tan»=2j;-l, tan2r= 'I 

39. y = 3t>«-4« + 6, t = 2x»-5. ^ = 72x6 -204x». 

dx 

-^ 2d X dy 4: 



8«-2 2x-l dx (x-2)« 

dy 
41. y = log (a^ — «^, v = asinx. ^ = ^2 tan x. 



42. y = arc tan (a + «),« = e^. 



dx 

dy e* 



dx 1 + (a + c»)* 

43. r = ^» + c», « = log(«-««). ^ = 4«»-6ea + l. 

44. to = 4 log -^ ^ -arc tan -— » 

y/l + Sz-\-Sz^ dw 1 



t) = 



dz «c (1 + z) 



* As was pointed out on p. 67, it might be possible to eliminate v between the two given 
expressions so as to find y directly as a function of jr, but in most cases the above method is to 
be preferred. 



RULES FOR DIFFERENTIATING 8S 

dx 
In the following examples first find — by differentiation and then snhstitute iu 

dy 





dz"^ dz 


to find ^^. 
dz 


dy 


46. x = yVl + y. 





46. x=vrT 



cosy. 



dy 


2 VI +y _ 


2x 




dz 


2 + Sy 2| 


r + 3y* 

2 




dy_ 


2 Vl + cos y 
siny 




dz 


%^ 


jP« 



47. x = -JL_. ^^ (l + logv)« 

1 + log y dx log y 



^o 1 « + Va» — y" dy y 

48. X = a log — . i' _ » 



dx ^Qi^^ 



49. x = rare vers?- V2;:ir^. *? = ^/?IEi. 

r dx \ y 

60. r = — ^ gr^^t^-^). 

1 + logA d« «> 



Ki 1 «* + Ve««'-4 

61. u = log — ! . 

* 2 

62. Show that the geometrical significance of XXVII is that the tangent makes 
complementary angles with the two codrdinate axes. 

75. Implicit functions. When a relation between x and y is 
given by means of an equation not solved for y, then y is called an 
implicit function of x. For example, the equation 

ar»-4y=:0 

defines y as an implicit function of x. Evidently x is also defined 
by means of this equation as an implicit function of y. Similarly 

x' + f + z'-.a!' = 

defines any one of the three variables as an implicit function of 
the other two. 

It is sometimes possible to solve the equation defining an implicit 
function for one of the variables and thus change it into an explicit 
function. For instance, the above two implicit functions may be 

solved for y giving 

or 

y=4 



and y=±Va^-ar^-«*; 



84 DIFFERENTIAL CALCULUS 

the first showing y as an explicit function of x^ and the second as an 
explicit function of x and z. In a given case, however, such a solu- 
tion may be either impossible or too complicated for convenient use. 
The two implicit functions used in this article for illustration 
may be respectively denoted by 

and F{^^ y, z) — 0. 

76. Differentiation of implicit functions. When y is defined as 
an implicit function of x by means of an equation in the form 

it was explained in the last section how it might be inconvenient to 
solve for y in terms of x\ that is, to find y as an explicit function of 
X so that the formulas we have deduced in this chapter may be ap- 
plied directly. Such, for instance, would be the case for the equation 

{B) aa* -f 2 ^y-y'x-l^ = 0. 

We then follow the rule : 

IHfferentiatey regarding y as a function of x^ and put the result 
equal to zero.* That is, 

' {€) ^/(x, y) = O. 

Let us apply this rule in finding -^ from {B). 

^Jaa? + 22^y-y''x-l(i) = 0; by (C) 

dx dx 

(2 2^ - 7 x/)^ = y^ - 6 arr^ - 6 2;^y ; 

dx 



Ans, 



dy ^y"^ — % aal^ — 6 u^y 
dx~ 2 7?^lxf 

The student should observe that in general the result will con- 
tain both x and y, 

* Tills procem will be justified in % 138, p. 202. Only corresponding values of x and y which 
satisfy the given equation may he substituted In the derivative. 



KULES FOR DIFFERENTIATING 85 



EXAMPLES 



Differentiate the following by the above rule. 

dy 2p 



1. y^ = ^px. 



2. za + y2 = T«. 



3. 62a;a4.aV = a2 6*. 



4. y8-3y + 2ax=:0. 



5. x^ + y* = a*. 



a y«-2a;y + 6« = 0. 



9. x« + y' — 3 oxy = 0. 



10. X9z=y^. 



11. p2 = a«C082^. 



12. pScos0=:a38in39. 



13. COS (uv) = cv. 



14. = cob(0 + 0). 



dx V 



dy_ 
dx 


X 

— — • 
V 


dy 
dx 


b»x 
a^y 


dy 


2a 


dx 


3(1 -y«) 


dy 
dx 


-4 



dx ^x 



dx 




a^ 


dy 
dx 


V 
V- 


X 


dy 


ay- 


-x« 


dx 


ya- 


- ox 


dy 


y^- 


- ay log y 


dx 


xa- 


- xy log X 


d^ 


a^ sin 2 
P 


dp 


8a9 


cos 3 ^ + p8 sin tf 


de 




2pCOB0 


du 


c + 11 sin (uv) 


dv 


— 1 


D sin (uv) 


de 




sin (9 + 0) 



d0 1 + sin (0 + ^) 



CHAPTER VII 

SIMPLE APPLICATIONS OF THE DERIYATIVE 

77. Direction of a cttnre. It was shown in § 45, p. 44, that if 

y =/(^) 
is the equation of a curve (see figure), then 




dy 
dx 



= tan T = slope of line tangrent to curve at any point P. 



The direction of a curve at any point is defined to be the same 
as the direction of the line tangent to the curve at that point. 
From this it follows at once that 



dy 
dx 



= tan T = slope of curve at any point P. 



At a particular point whose coordinates are known we write 






slope of curve (or tangent) at point (xi^ f^i). 



y=Vx 



At points such as 2>, Fy JJ, where the curve (or tangent) is 
parallel to the axis of JT, 

T = 0% therefore ^ = O. 

ax 

At points such as Ay B, Gj where the curve (or tangent) is 
perpendicular to the axis of X^ 

T = 90% therefore ^ = «. 

ax 

86 



SIMPLE APPLICATIONS OF THE DERIVATIVE 



87 



At points such as E^ where the curve is rising^* 

a positive number. 



T = an acute ansrie, therefore -^ = 

ax 




The curve (or tangent) has a positive slope to the left of B^ 
between D and Fy and to the right of G. 

At points such as (7, where the curve isfallinffy* 

T = an obtuse ansrle, therefore —^ = a negative number. 

The curve (or tangent) has a negative slope between B and 2> 
and between F and G, 

Ex. 1. Given the curve y = — — x* + 2 (see flgore). 

3 

(a) Find t when x = 1. 

(b) Find r when x = S, 

(c) Find the points where the curve is parallel to OX. 

(d) Find the points where r = 45°. 

(e) Find the points where the curve is parallel to the line 
2x-3y = 6 (line -4^. 

dy 
Solution. Differentiating, — = 2b> — 2 x = slope at any point. 

dx 

(a)tonT=r^l = 1 -2 = -1; therefore T = 186° Ans. 

(b) tan T=-~ =9 — 6 = 8: therefore t = arc tan 8. Ans, 

Ldxj Xa 3 

(c) T = 0°, tan T = — = ; therefore x" — 2 x = 0. Solving this equation, we And 

dx 

that X = or 2, giving points C and D where curve (or tangent) is parallel to OX. 

(d) T = 45°, tan T = — = 1: therefore x« - 2x = 1. Solving, we get x = 1 ± V2, 

dx 

giving two points where the slope of curve (or tangent) is unity. 

(e) Slope of line = | ; therefore x* — 2 x = f . Solving, we get x = 1 ± Vj, giving 
points E and F where curve (or tangent) is parallel to line AB. 

Since a curve at any point has the same direction as its tangent 
at that point, the angle between two curves at a common point 
will be the angle between their tangents at that point. 

Ex. 2. Find the angle of intersection of the circles 
{A) x2 + ya-4x = l, 

(B) X» + y*-2y = 9. 



* When moving from left to right on carve. 



88 



DIFFERENTIAL CALCULUS 



Solutu 
and (1, - 

y 


m. Solving simultaneously we find the points of intersection to be (3, 2) 
-2). 

dx y 


r 


^^ 




**"= * from(B). 


By § 76, p. 84 


W^ 


"2 


= — i = slope of tangent to 

y Jx=8 

^'2 


(A) at (3, 2). 


j:> 


^rr^ 


r_ 


1 = — S = nInnA nf tAncrAnt: to 


ITC\ at l^ 2^ 



|f=2 

The formula for finding the angle between two lines whose slopes are mi and m^ is 

fill — Wlj 



tan^ = 
« 1 + w,\in% 

— i 4- 3 
Substituting, tan ^ = — UL- = i ; therefore 6 = 45°. Aia, 

This is also the angle of intersection at point (1, — 2). 



65, p. 3 



EXAMPLES 
The corresponding figure should be drawn in each of the following examples. 



1. Find the slope oiy — 



1 + xa 



at the origin. 



A%i», 1 = tan r. 



2. What angle does the tangent to the curve x^^ = a^ (x + y) at the origin 
make with the axis of X? Aw&. r ^ 135°. 

3. What is the direction in which the point generating the graph of y = 3 x^ — x 

tends to move at the instant when x = 1 ? 

^n«. Parallel to a line whose slope is 6. 

4. Show that — (or slope) is constant for a straight line. 

dx 

5. Find the points where the curve y = x' — 3x* — 9x + 5 is parallel to the 
axis of X. Am. X = 3, X = — 1. 

6. Fiud the poijits where the curve y (x — 1) (x — 2) = x — 3 is parallel to the 
axis of X, Aivi, x = 3 ± V^. 



7. At what point on j/^ = 2 x^ is the slope equal to 3 ? 



Am. (2,4). 



8. At what points on the circle x^ + y' = r^ is the slope of tangent line equal 



to -f ? 



A-M 



■ ( 



3r 



5 / 



9. Where is the tangent to the parabola y = x>-7x + 3 parallel to the line 
y = 6x + 2? Aim. (6, -3). 

10. Find the points where the tangent to the circle x^ + y^ = 169 is perpen- 
dicular to the line 5x + 12 y = 60. Am. (±12, T 5). 



SIMPLE APPLICATIONS OF THE DERIVATIVE 89 

11. Find the point where the tangent to the parabola ^ = 4 ox is parallel to the 
Jine x-\- y = 2, ' Ans, (o, — 2 a). 

12. At what angles does the line 3 y - 2 x — 8 = cut the parabola y^ = Sx. 

Am. arc tan \^ and arc tan \, 

13. Find the angle of intersection between the parabola ^^ = 6 x and the circle 
x2 + y2 _ 16, ^^^ arc tan f Vs. 

X^ «/2 

14. Show that the hyperbola x^ -- y* = 5 and the ellipse f- — = 1 intersect at 

right angles. 

x' 

15. Show that the circle x' + ^ = 8 ox and the cissoid v' = 



2a — X 

(a) are perpendicular at the origin ; 

(b) intersect at an angle of 45° at two other points. 

16. Find the angle of intersection of the parabola x^ = 4 ay and the witch 

8a' 
y = Am. arc tan 3 = 71° 33'. 9. 

x* + 4 a* 

17. Show that the tangents to the folium of Descartes x' + y^ = 3 axy at the 
points where it meets the parabola y^ = ox are parallel to the axis of F. 

18. At how many points can the curve y = x' — 2x2 + x — 4 be parallel to the 
axis of X? What are the points ? An». Two ; at (1, - 4) and (\, - V^). 

. 19. Find the angle at which the parabolas y = 3 x^ — 1 and y = 2 x^ + 3 intersect. 

Am. arc tan ^. 

20. Find the relation between the coefScients of the conies aix' + 5iy^ = 1 and 
Oax* + ha/^ = 1 when they intersect at right angles. . 1_1_1^1 

ai 6i Oa 6i 

78. Equations of tangent and normal, lengths of subtangent and 
subnormal. Rectangular coordinates. The equation of a straight 
line passing through the point (a^^, y^ and having the slope m is 

y-y^^m{x- x^. 64 (c), p. 3 

If this line is tangent to the curve AB at 
the point Pi{x^y y^), then from § 77, p. 86, 

_dy* 



= tanT = r$?l =1^^ 
\_dxj :r= xi dx^ 




M N X 



m 

\__dxj 

V-Vl 

Hence at point of contact Py (orj, y^ the equation of the tangent 
line TP^ is 

• By this notation is meant that w© should first find -^, then in the result substitute x-^ for x 

(Ix J 

and w, for y. The student is warned against interpreting the symbol — to mean the (ierira- 

rt.r, 

tive o/ffi teieh respect to x^ for that has no meaning whatever since Xj and f/i are both constants. 




90 DIFFERENTIAL CALCULUS 

The normal being perpendicular to tangent, its slope is 

-i=-S- By65,p.3 

And since it also passes through the point 
of contact Fy^{x^y y^), we have for the equation 
M N X of the normal P^N 

That portion of the tangent which is intercepted between the 

point of contact and OX is called the length of the tangent (= TP^^ 

and its projection on the axis of X is called the length of the 

subtangent (= TM). Similarly we have the length of the normal 

(= P^N) and the length of the subnormal (= MN), 

MP 
, In triangle TP^My tan t = -— i ; therefore 

MP dxi 

(8) TM* = = yi -^ = lengrth of subtangrent. 

^ ' tan T dyi 

MN 
In the triangle MP^N^ tan t = -rr^; therefore 

(4) MN\ = MP^ tan T = yi ^ = lengrth of subnormal. 

The length of tangent (== TP^ and the length of normal (= P^N) 
may then be found directly from the figure, each being the hypote- 
nuse of a right triangle having the two legs known. Thus 

(5) = y 1 -%/ ^ — i ^ + 1 = lengrth of tangrent. 



P,N = y/MP^ + & = ^(y,)» + (y, ^ J 



(6) = yi -Jl + ( ^ )* = lengrth of normaL 

The student is advised to get the lengths of the tangent and of 
the normal directly from the figure rather than by using (5) and (6) 
as formulas. 

* If Bubtangent extends to the right of T, we consider it positire ; if to the left, negative, 
t If subnormal extends to the right of My we consider It positlre ; if to the left, negative. 



SIMPLE APPLICATIONS OF THE DERIVATIVE 



91 



EXAMPLES 

1. Pind the equations of tangent and normal, lengths of subtangent, sabnonnal 
tangent, and normal at the pomt (a, a) on the cissoid jfi = 



2a — z 



Solution. 



dy __ 3 ax^ — afi 
dx~ y(2a-x)^ 
dj/i rdyl 3 a* — a" 



Hence ^> = ^ = _3^i!ji:*L = 2 = dope. 
dzi LdxJx^a a(2a-a)« 

Sabstituting in (1) gives 

y = 2x — a, equation of tangent. 
Substituting in (2) gives 

2 y + X = 3 a, equation of normal. 

Substituting in (3) gives 

d 
TM = - = length of subtangent. 

Substituting in (4) gives 

MN = 2 a = length of subnormal. 




Also, FT = V( TMf + (ilfP)2 = -y/— + a^ = ? Vs = length of tangent, 

and PN = V(3fiV^)2 4. (MP)^ = vToM^ = a Vs = length of normal. 

2. Find equations of tangent and normal to the ellipse x^ -^ 2y^ — 2xy — z = 
at the points v^here x = 1. Ara. At (1, 0), 2 y = x — 1, y + 2 x = 2. 

At (1, 1), 2y = x+ 1, y + 2x = 3. 



3. Find equations of the tangent and normal, lengths of subtangent and sub- 
normal at the point (Xi, yi) on the circle x^ + y^ = r^. ^ 

Ans. xix 4- yiy = r^, Xiy — yjx = 0, — — 1 Xi. 

Xi 



4. Show that the subtangent to the parabola y' = 4 px is bisected at the vertex, 
and that the subnormal is constant and equal to 2 p. 

5. Find the equation of tangent at (xi, yi) 

(a) to the ellipse ^ + ^- =1 ; (b) to the hyperbola — - ^ = 1. 



a* 



63 



63 



6. Find equations of tangent and normal to the witch y = 
point where x = 2 a. 



8a« 



at the 



4 o^ -f x2 
Ans, x + 2y = 4a, y = 2x--3a. 



_ at __» 

7. Prove that at any point on the catenary y =. -{(^ ■\- e *") the lengths o\ 

subnormal and normal are - (6 " — e <> ) and ^ respectively. 

4 a 



92 DIFFERENTIAL CALCULUS 

8. Find the equation of tangent to the conic ox^ + 2bxy + cy* + 2(Ix-f2ey 
+/= at the point (xi, yi). 

Ans. axix + b (yix + xiy) 4- cyiy + d(xi + x) + e{yi + y) +/= 0.» 

9. Show that the equation of tangent to cunre (-j +(^) =2 at the point 

(a, &) is - + - = 2 for all values of n. 
a b 

10. Prove that the length of subtangent toy = O'Ib constant and equal to 



log a 

11. Get the equation of tangent to the parabola y^ = 20x which makes an angle 
of 46° with the axis of x. Ans, y = x + 5. 

Hint. FInt find point of contact by method of Ex. 1 (e), p. 87. 

12. Find equations of tangents to the circle x^ + ^ = 62 which are parallel to 
the'line2x + 3y = 6. Ara. 2x + 3y±26 = 0. 

13. Find equations of tangents to the hyperbola 4x9 — 9 y* + 36 = which are 
perpendicular to the line 22^ + 6x = 10. Ans. 2X'- 6y ±S = 0. 

i4. Show that in the equilateral hyperbola 2 xy = a^ the area of the triangle 
formed by a tangent and the codrdinate axes is constant and equal to a^. 

15. Find equations of tangents and normals to the curve y^ = 2x^ — x* at the 
points where x = 1. Ans. At (1, 1), 2 y = x + 1, y 4- 2 x = 3. 

At (1, -1), 2y = -x-l, y-2x = -3. 

16. Show that the sum of the intercepts of the tangent to the parabola 

X* -f y* = a* 
on the cobrdinate axes is constant and equal to a. 

17. Find the equation of tangent to the curve x* (x + y) = a* (x — y) at the origin. 

Ans. y = X. 

18. Show that for the hypocycloid x' + y' = a' that portion of the tangent 
included between the coordinate axes is constant and equal to a. 

X 

19. Show that the curve y = ae^ has a constant subtangent. 

20. Show that the length of tangent is constant in the tractrix 

X = vc2 - ya + - log ^ 



2 c+V^^^^ 

79. Parametric equations of a cunre. Let the equation of a 
curve be 

(A) F{x,y)=0. 

If X is given as a function of a third variable, a say, called a 
parameter^ then by virtue of (A) y is also a function of a, and 

* In Ex8. 3, 6, and 8 the student should notice that if we drop the tubecrlpts in equations of 
tangents they reduce to the equations of the curves themselves. 



SIMPLE APPLICATIONS OF THE DERIVATIVE 



93 



(C) 




the same functional relation (A) between x and y may generally be 
expressed by means of equations in the form 

(x =/(a), 
(^ \y=<t>(a}i 

each value of a giving a value of x and a value of y. Equations 
(B) are called parametric eqiuitions of the curve. If we eliminate 
a between equations (-B), it is evident that the relation (A) must 
result. For example, take equation of circle 

a? H- y* = r*, or y = Vr' — 2?, 

Let 2;=rcosa; then 

y = r sin a, and we have 
'x=r cos a, 
j/=r sin a, 

as parametric equations of the circle in figure, 
« being the parameter. 

If we eliminate a between equations (C) by squaring and adding 
the results, we have 

2^ -{-if^ =z r^(cos^ a -f- sin' a) = r^, 

the rectangular equation of the circle. It is evident that if a 
varies from to 27r, the point -P(a-, y) will describe a complete 
circumference. 

In § 84, p. 104, we shall discuss the motion of a point P, which 
motion is defined by equations such as 

\y = (^(«). 

We call these the parametric equations of the path, the time t 
being the parameter. Thus in Ex. 2, p. 106, we see that 

Xz=Vq cos a • ^, 

l/ = -i^fft^ -\-v^sin a-t 

are really the parametric equations of the trajectory of a projectile, 
the time t being the parameter. The elimination of t gives the 
rectangular equation of the trajectory 



y = a: tan a — 



ff^ 



2 Vq cos'^ a 



94 



DIFFERENTIAL CALCULUS 



Since from (B) y is given as a function of a^ and a as a function 
of iT, we have 

dy dy da 
dx da dx 



by XXVI 



da dx 



by XXVII 



da 



that is, 
(1» 



dy 
dy _ da 
dx " doc 

da 



Hence, if parametric equations of a curve are given, we can find 
equations of tangent and normal, lengths of subtangent and sub- 
normal at a given point on the curve, by first finding the value 

of -~ at that point from (I>) and then substituting in formulas 

ClX 

(1), (a), (8), (4) of last section. 



Ex. 1. Find equations of tangent and normal, lengths of subtangent and sub- 
normal to the ellipse 



(E) 



at the point where 4> = - 

4 



!x = a cos 0, 
y = 6 sin 0,* 



dz 



Solution. The parameter being 0, -- = — a sin 0, -=^ = 6 cos 0. 

cUp a0 

Substituting in (D), -=^ = :— 5 = slope at any point. 

dx asin <f> 



*A8 in figure draw the major and minor anziliary 

droles of tlie ellipse. Throagh two points B and C on the 

same radius draw lines parallel to the axes of codrdinates. 

These lines will intersect in a point P (x, y) on the ellipse, 

because 

a:= OA = OB cos ^ = a cos ^ 

and y ^ AP= OD=OC Bin it^b^n^f 



Y 


^ 


3 


/^y^"^^^ 


/4 




11 ^ 


J 


yj 



or, 



cr t/ 

-=GOs6 and -=sinA. 
a ^ b 



Now squaring and adding we get 

— + ?^=coB«0 + Bin«<^=l, 
a* b* 

the rectangular equation of the ellipse. ^ Is sometimes 
oalled the eccentric angle of the ellipse. 



SIMPLE APPLICATIONS OF THE DERIVATIVE 96 

as the point of 



T . 



Sabstituting = - iu the given equations {E)^ we get 

4 



contact Hence 



Sabstituting in (1), p. 80, 



(v^' V2) 



dVi 
dxi 



h 
a 



y- 



V2 



=-^(*-^> 



or. 



Substituting in (2), p. 00, 



or. 



bx + a]^ = V2 a&, equation of tangent 
^ _ ? / _ q \ 

V2 (ox — 6y) = a" — 62, equation of normal. 



Substituting in (3) and (4), p. 00, 

— ( — ^ = ^ = length of subnormal. 

'2 V a/ aV2 



62 



V2 
b 



^ = = length of subtangent 

_ M V2 



Ex. 2. Given equation of the cycloid* in parametric form 

ix = a(5 — sin 6), 
y = a (1 — cos 6) ; 

6 being the variable parameter. Find lengths of subtangent, subnormal, tangent, 

and normal at the point where ^ = - • 

2 



Solution, 



dx dy 

— = a(l — cos ^, -^ = o sin ^. 

de ^ ' de 



*The path described by<a point on the droomference of a circle which rolli without sliding 
on a fixed straight line is called the cycloid. Let the radius of rolling circle be a, /> the gener- 
ating point, and M the point of contact with the fixed line OJT, which is called the base. If arc 




PM equals OM in length, then P will touch at O if circle is rolled to the left. We hare, denoting 

angle PCM by 0, 

a: = OJf - J^^= o« - a sin « = o(« - sin tf ) , 

y= P^= AfC- -4C=a- acos tf=a(l- cos ») ; 

the parametric equations of the cycloid, the angle B through which the rolling circle turns 
being the parameter. 0D^2 ira is called the base of one arch of the cycloid, and the point V 
is called the yertez. Eliminating 9, we get the rectangular equation 



x^aarcoosT — ^j- V2ay->y>. 



96 DIFFERENTIAL CALCULUS 

Substituting in (D), p. 94, ~ = = slope at any point. 

dx 1 — cos d 

Since 5 = - 1 the point of contact is ( a, a V and -— = 1. 

2 \ 2 / dxi 

Substituting in (3), (4), (6), (6) of last section, we get 
length of subtangent = a, 
length of subnormal = a, 
length of tangent = a \^, 
length of normal = a V^. Aim. 

Note. Draw the tangent PT, the vertical diameter MB^ and 

connect P and B. 

^ , 
, ^ 2 sin - cos - ^ 

X ^rn.r^ ^V sin ^ 2 2 

tan MTF = -^ = = — = cot -• 

dx 1-cos^ ^ . ^0 2 

2 sin^ - 

[From 37, p. 2, and 39, p. 3.] 

Hence angle MTF = ^ - h ' By 29, p. 2 

Q 

Ako, angle PBM= -, since it is measured by one half 

the arc MP which measures the central angle 0^ and we have 

ansfle APB = — — - . 
^ 2 2 

Comparing, we see that 

angle MTP = angle APB. 

Therefore : 

The tangent to a cycloid always passes through the highest point 
of the generating circle. 



EXAMPLES 

In the follo'wing curves find lengths of (a) subtangent, (b) subnormal, (c) tangent, 
(d) normal, at any point. 



^ rm. {x = a (cos ^ + ^ sin t), 

1. The curve \ , • , ^ / 



Ana, (a) ycoitj (b) ytant, 



sin t cos t 



SIMPLE APPLICATIONS OF THE DERIVATIVE 97 



2. The hypocycloid (astroid) 



X = 4 a 006* t, 
y = ia sin' t 

Ana, (a) — y cot t, (b) — y tan t. 




sin t cos t 

« r»«- J- .J (« = a(2co8t - C08 2Q, 

3. Thecardioid { /o • * • o/ 

( y = a (2 sin t — sin 2 1). 

80. Angle between the radius vector drawm to a point on a curve 
and the tangent to the curve at that point. Let the equation of 
curve in polar coordinates be p =/(^). 

Let F be any fixed point (/>, 0) on the cui-ve. If 0y which we 
assume as tbe independent variable, takes 
on an increment Ad, then p will take on a 
corresponding increment A/). Denote by 
Q the point (p + A/), d^ A0). Draw PE 
perpendicular to OQ. Then OQ^zp-^^Ap^ o 
PR = p sin Ad, and OR = p cos Ad. Also, 

. _,__ PR PR p sin Ad 

tan POR = = = ::• 

^ RQ OQ-OR /) -f- A/3 - /3 cos Ad 

Denote by -^ the angle between the radius vector OP and the 
tangent PT, If we now let Ad approach the limit zero, then 

(a) the point Q will approach indefinitely near P; 

(b) the secant PQ will approach the tangent PT as a limiting posi- 
tion; and 

{c) the angle PQR will approach ylr as a limit. 

Ad = /) -f A/) — /» COS Ad 

^ limit P sin Ad 

Ad = _ . ,Ad ^ 
zpsur— -f A/» 

j since from 39, p. 3, p-p oos A*- p (1 - OM A*)- 2p lin*-^..] 

^ limit Ag 

A^ = 0„ . ,A^ 

^^^ Ap 

[Diyiding both numerator and denominator by A0.] 



98 



DIFFERENTIAL CALCULUS 



limit 



P- 



sin Ad 



. Ad 
. Ad '"'T Ap 



c;„«^ limit /^P\ ''P „„a limit / •„ ^^\ a i 
^^"^'^ Ad = o(^Ai; = i ^"'^ Ad = o("°TJ = ®' *^ 



limit / sin Ag \ 
Le = (S\ A0 J 



(-4) 



. A0 
sin-— 

= land^li=^\ L: 



A^ 
2 



= 1 by (14), p. 30, we have 






From the triangle OFT we get 
(JB) r = e + tlr. 

Ex. 1. Find ^ and r in the cardioid p = a (1 — cob 9). 

dp 
Solvtion, -^ = a sin 9. Substituting in (^) gives 

2 a sin2 - 
p a(l-cos^) 2 

tan^ = -^ = — ^^ = ^^^— ^-^— 

dp aame ^ . 




dS 



2 a sin - cos 
2 2 



= tan -. By 89, p. 3, and 37, p. 2 



6 0^0 

Since tanf = tan-i ^ = -. Ana. Substituting in (B), t = ^ + - = — Ans, 

A Jt 2 2 



81. Lengths of polar subtangent and polar sub- 
normal. Draw a line NT through the origin pei^ 
pendicular to the radius vector of the point P on 
the curve. If FT is the tangent and FN the 
normal to the curve at P, then 




and 

of the curve at P. 



OT = length of polar subtangent^ 
0N= length of polar subnormal 



SIMPLE APPLICATIONS OF THE DERIVATIVE 99 

OT 
In the triangle OPT^ tan -^ = Therefore 

P 

(7) or = /> tan-^ = /)*— = len^h of polar subtangent.* 

up 

In the triangle OFN^ tan -^ = -^. Therefore 

(8) ON^ — ^-- = — ^ = lencrth of polar subnormaL 

tan'^ »^ 

The length of the polar tangent (= FT) and the length of the 
polar normal (= PN) may be found from the figure, each being 
the hypotenuse of a right triangle. 

Ex. 1. Find lengths of polar subtangent and sabnormal to the lemniscate 
f^ = a^ cos 2 e. 

Solution, Differentiating as an implicit function with respect to 6, 

2p^ = -2a»sin2(^, or*? = -?^^l^^ 

'^de de p 

Snhstitating in (7) and (8), we get 

p* 
length of polar subtangent = ^ 1 

* *^ ' * a«sin2^ 

length of polar sabnormal = • 

P 

If we wish to express the results in terms of 6^ find p in terms of 6 from the given 

equation and substitute. Thus , in ab ove, p = ± a Vcos 2 ; therefore length of 

polar subtangent = ± a cot 2 ^ Vcos 2 6, 



EXAMPLES 

1. In the circle p = r sin 9, find rp and r in terms of 0. Ana. ^ = ^, r = 2 ^. 

2. In the parabola p = a sec^ -i show that r + ^ = ir. 

3. Shpw that ^ is constant in the logarithmic spiral p = e^'. - Since the tangent 
makes a constant angle with the radius vector this curve is also called the equiangular 
spiral. 

$ 

4. Given the conchoid p =:a sin* - ; prove that r = 4 ^. 

o 

6. Show that tan ^ = ^ in the spiral of Archimedes p = a0. Find values of \f/ 
when^ = 2ir and4ir. Ans, f = 80° 57' and 85*» 27'. 

• When » increaaes with p, — is positive and i^ is an acnte angle, as lu above figure. Then the 

dp 

subtangent OTls positive and is measured to the right of an observer placed at O and looking 
along OP. When — is negative the subtangent is negative and is measured to the left of the 
observer. 



100 DIFFERENTIAL CALCULUS 

6. Find ^ in the curves p* = a» sin nd and p» = 6" cos n$, 

Ans. rp =:n0 and „ + n^. 

7. Show tliat the curves in the preceding example intersect at right angles. 
Hint, Find r for each curve and compare. 

8. Prove that the spiral of Archimedes p = a0t and the reciprocal spiral p = - , 
intersect at right angles. 

9. Find the angle between the parabola p = a sec' - and the straight line 
p sin ^ = 2 a. Ara. 45°. 

10. Show that the two cardioids p = a (1 + cos ^) and p = a (1 — cos 0) cut each 
other perpendicularly. 

11. Find lengths of subtangent, subnormal, tangent, and normal of the spiral of 

Archimedes p = a^. ^ u* P^ * P ^nr, — ? 

'^ Ans. subt. = — , tan. = - va^ + p2, 

a a 

subn. = a, nor. = Vcfi~+~^, 
The student should note the fact that the subnormal is constant. 

12. Get lengths of subtangent, subnormal, tangent, and normal in the logarithmic 

spiral p=a9. p I j 

Ana. subt. = ——, tan. = p-\/l + ; — r— » 

log a \ log2 a 

subn. = p log a, nor. ='p Vl -f log* a. 

When a = 6 we notice that subt = subn., and tan. = nor. 

13. Find the angles between the curves p = a(l + cos 5), p = 6(1 — cos $). 

Ana. and - • 
2 

14. Show that the reciprocal spiral p = - has a constant subtangent. 



15. Prove that the curves p» = a* co8(nd — a) and p^ =ia* cos {n0 — /3) Intersect 
at an angle a — j9. 

16. Show that the area of the circumscribed square about the cardloid 

p = a(l — cos^) 

formed by tangents inclined 46° to the axis is f { (2 + Vs)a^. i 

82. Solution of equations having multiple roots. Any root which 
occurs more than once in an equation is called a multiple root. 
Thus 3, 3, 3, — 2 are the roots of 

(A) a:* - 7 a:* + 9 a:* + 27 a: - 54 = ; 

hence 3 is a multiple root occurring three times. 
Evidently (A) may also be written in the form 

(a; - 3)« (a: 4- 2) = 0. 



SIMPLE APPLICATIONS OF THE DERIVATIVE 101 

Let f{x) denote an integral rational function of x having a mul- 
tiple root a, and suppose it occurs m times. Then we may write 
{B) f{x)=={x^ar<f>{x), 

where <l>{x) is the product of the factors corresponding to all the 
roots of f{x) differing from a. Differentiating (5), 

/ (x) = (a; - a)-* <^' (x) + <^ (2:) m (rr - a)^-\ or, 
(C) f{x) = {x- a)-> [{X - a) 4>' (X) + <^ (x) m]. 

Therefore /'(a:) contains the factors (x — a) repeated w — 1 times 
and no more ; that is, the highest common factor (H.C.F.) of f{x) 
and f (x) has w — 1 roots equal to a. 

In case f{x) has a second multiple root fi occurring r times, it is 
evident that the H.C.F. would also contain the factor (a: — ^8)'""'*, 
and so on for any number of different multiple roots, each occur- 
ring once more in /(a:) than in the H.C.F. 

We may then state a rule for finding the multiple roots of an 
equation f{x) = as follows : 

First step. Find f (x). 

Second step. Find the K C.F, off(x) and f {x). 
Third step. Find the roots of the H.CF. Each different root 
of the H.C.F. will occur once more inf{x) than it does in the H.C.F. 

If it turns out that the H.C.F. does not involve x^ then/(a:) has 
no multiple roots and the above process is of no assistance in the 
solution of the equation, but it may be of interest to know that 
the equation has no equaU i-e. multiple^ roots. 

Ex. 1. Solve the equation «» - 8x* + 13x -6 = 0. 
Solulum. Place f{z) = x* - 8 «» + 13 x - 6. 

First step. f(z) = 3 x^ ~ 16 x + 13. 

Second step. H.C.F. = x — 1. 

Third step. x - 1 = 0. .-. x = 1. 

Since 1 occurs once as a root in the H.C.F. it will occur twice in the given equa- 
tion ; that is, (x - 1)' will occur there as a factor. Dividing x" — 8x* + ISx — 6 
by (x — 1)2 gives the only remaining factor (x — 6), yielding y 
the root 6. The roots of our equation are then 1, 1, 6. 
Drawing the graph of the function, we see that at the double 
root X = 1 the graph touches OX but does not cross it.* 

*Sinee the flrat deriyative yanishefl for every multiple root, it 
follows that the axis of X is tangent to the graph at all points corre- 
sponding to multiple roots. If a multiple root occurs an even number 
of times, the graph will not cross the axis of X at such a point (see 
figure) ; If it occurs an odd number of times, the graph will cross. 




102 DIFFERENTIAL CALCULUS 

EXAMPLES 

Solve the first ten equations by the method of this section. 

1. x>-7x« + 16x-12 = 0. Am. 2,2,3. 

2. z*-6x2-8x-3 = 0. Am, -1,-1,-1,8. 

3. x*-7x« + 9x2 + 27x-54 = 0. Aub. 3, 3, 3, - 2. 

4. x*-6x«-9x« + Six -108 = 0. Ans, 3,3,3,-4. 
6. x* + 0x" + x2-24x + 16 = 0. Am. 1,1,-4,-4. 

6. X* - 9x» + 23x2 -3x- 36 = 0. Am. 3,3,-1,4. 

7. x*-6x» + 10x2-8 = 0. Am. 2, 2, 1±V3. 

8. x»-x*-6x« + x2 + 8x + 4 = 0. Am. -1,-1,-1,2,2. 

9. x* - 16x« + 10x2 + 60x - 72 = 0. Am. 2, 2, 2, - 3, - 3. 

10. x6 - 3x* - 5x» + 13x2 + 24X + 10 = 0. Am. - 1, - 1, -1, 3 ± VTl. 

Show that the following four equations have no multiple (equal) roots. 

11. x» + 9x2 + 2x- 48 = 0. 

12. X* - 15x2 -10x + 24 = 0. 

13. X* - 3x« - 6x2 + i4x + 12 = 0. 

14. x»-a» = 0. 

16. Show that the condition that the equation 

x» + 3g'x + r = 
shall have a double root is 4 g* 4- r2 = 0. 

16. Show that the condition that the equation 

x> + 3i)x2 + r = 
shall have a double root is r (4p8 -^ r) = o. 

83. Applications of the derivative in mechanics. Velocity. Con- 
sider the motion of a point P describing a 
""^ curve AB, Let « be the distance measured 

y 

along its path from some fixed point as A 

to any position of P, and let t be the corre- 

— X sponding elapsed time. To each value of t 

corresponds a position of P in the path and 

therefore a distance (or space) «. Hence 8 will be a function of <, 

and we may write 



SIMPLE APPLICATIONS OF THE DERIVATIVE 103 

Now let t take on an increment A^ ; then. a takes on an increment 
A«, and 

{A) — = magnitude of the average velocity * 

of P during the time interval At. If F moves with uniform motion, 
the above ratio will have the same value for every interval of time 
and is the speed (= magnitude of the velocity) at any instant. 

For the general case of any kind of motion, uniform or not, we 

define the speed v (= magnitude of the velocity) at any instant as 

A« 
the limit of the ratio — - as At approaches the limit zero ; that is. 



At 



.. limit A« _ 

v= A^ A "r-» or, 

Ae = At 



The speed (= magnitude of the velocity) for afiy motion is the deriv- 
(Uive of the distance (= space) with respect to the time. 

To show that this agrees with the conception we already have 
of speed, let us find the speed (= magnitude of the velocity) of a 
falling body at the end of two seconds. 

By experiment it has been found that a body falling freely from rest 
in a vacuum near the eailh's surface f oUows approximately the law 

{B) s = 16.1 e, 

where s = space fallen in feet, t = time in seconds. Apply the 
General Rule^ p. 42, to {B). 

First step. «4.A«= 16.1 (t + At)* = 16.1 t»-f. 32.2 <• At +16.1 (Ae)». 
Second step. A« = 32.2 1 . At + 16.1 (At)^ 

A8 

Third step. — = 32.2 1 -f- 16.1 At = average speed (= magnitude 

of the average velocity) throughout the time interval At reckoned from 

any fixed instant of time.^ 

Placing t = 2, 

A« 
{C) — = 64.4 4- 16.1 At = average speed throughout the 

time interval At aft^r two seconds of falling, 

* Velocity ia defined as the time rate of change of place, and is a rector quantity, 
t £a being the space or distance passed over In the time A<. 



104 DIFFERENTIAL CALCULUS 

Our notion of speed tells us at once that (C) does not give us 
the actual speed at the end of two seconds ; for even if we take At 
very small, say ^^^ or j-^-(^ of a second, (C) still gives only the 
average speed during the corresponding small interval of time. But 
what we do mean by the speed at the end of two seconds is the 
limit of the average speed when At diminishes towards zero; that is, 
the speed at the end of two seconds is, from (C), 64.4 ft. per second. 
Thus even the everyday notion of speed which we get from experi- 
ence involves the idea of a limit, or in our notation 

V = ^^ ( A^ ) ~ ^^'^ ^** P®^ second. 

The above example illustrates well the notion of a limiting value. 
The student should be impressed with the idea that a limiting value 
13 a definite^ fixed value, not something that is only approximated. 
Observe that it does not make any difference how small 16.1 A^ 
may be taken ; it is only the limiting value of 

64.4 4- 16.1 At, 

when A^ diminishes towards zero, that is of importance, and that 
value is exactly 64.4. 

84. Component velocities. The coordinates x and ^ of a point 
P moving in the XY plane are also functions of the time, and 
the motion may be defined by means of two equations, 

Tjl^^^- These are the parametric equations of 
Z/^^^ ^^ V^^ (see § 79, p. 92). 
■-"" f I The horizontal component v^ of t; f is the 

— ^ i velocity along OX of the projection J!f of P, 

^' and is therefore the time rate of change 

of X. Hence, from (9), p. 103, when s is replaced by x, we get 

(10) «^* = ^- 

at 

In the same way we get the vertical component, or time rate of 
change of y, 

(11) ..=f. 

* The equation of the path in rectangular coiirdinates may be found by eliminating t between 
these equations. 

t The direction of t; Is along the tangent to the path. 



SIMPLE APPLICATIONS OF THE DERIVATIVE 105 

Representing the velocity and its components by vectors, we 
have at once from the figure 

««> "=g=V(f)'+(f)"' 

giving the speed (= magnitude of velocity) at any instant. 

If T be the angle which the direction of the velocity makes with 
the axis of X^ we have from the figure, using (9), (10), (11), 

dy day dy 

(18) sinr = -^ = -7-; cost = — = -—; tanr = -^ = -t— 

V d« V dS^ Vao rf» 

dt dt dt 

85. Acceleration. In general v wil^ be a functipn of t and we 

Now let t take on an increment A^, then v takes on an increment 

Av, and 

Av 

— = magnitude of the average acceleration* of P during the time 

interval Af. 

For any kind of motion we define the magnitude of the accelera- 

Av 
tion a at any instant as the limit of the ratio — as A^ approaches 

the limit zero ; that is. 



__ limit /AvN 

Ae = o\^Ae/ 



^ = Ae -'--»'^^' 
dv 



The magnitude of the acceleration for any motion is the derivative 
of the velocity with respect to the time. 

86. Component accelerations. Following the same plan used in 
finding the component velocities, we get for the component accelera- 
tions parallel to OX and OF, 

,^-v dVac dVu A-i 



giving the magnitude of the acceleration. 

* Acceleration is defined as the time rate of change of Telocity, and is a vector quantity. 



106 DIFFERENTIAL CALCULUS 



EXAUPLSS 

1. By experiment it has been found that a body falling freely from rest in a 
vacuum near the earth's surface follows approximately the law 

d = 16.1<2, 

where s = space (height) in feet, t = time in seconds. Find magnitudes of the 
velocity and acceleration 

(a) at any instant ; 

(b) at end of the first second ; 

(c) at end of the fifth second. 

Solution, {A) «=16.ie>. 

ds 

(a) DifferentiaUng, — = 32.2 1, or, from (9), 

dt 

{B) V = 82.2 1 ft. per sec. 

/?« 
Differentiating again, — = 32.2, or, from (14), 

(O o ?= 32.2 ft. per (sec.)«, 

which tells us that the acceleration of a falling body is constant ; in other words, the 
velocity increases 82.2 ft. per sec. every second it keeps on falling. 

(b) To find V and a at the end of the first second, substitute t = 1 in (B) and (C) ; 

V = 32.2 ft. per sec., 
a =r 82.2 ft. per (sec.)^. 

(e) To find v and a at the end of the fifth second, substitute t = 6 in (B) and (C) ; ' 

v = 161 ft. per sec, 
a = 32.2 ft. per (sec.)«. 

2. Neglecting the resistance of the air, the equations of motion for a projectile are 

X = Vi cos <f> • t, 
y = Vism<p-t — 16.1fi; 

^^^ ^>^ where Vi = initial velocity, <p = angle of projection with 

^ ^v horizon, t = time of fiight in seconds, x and y being meas- 

A X ^^Li^d in feet. Find the magnitudes of velocity, accelera- 
tion, component velocities, component accelerations 

(a) at any instant ; 

(b) at the end of the first second, having given Vi = 100 ft. per sec., = 80° ; 

(c) find dii-ection of motion at the end of the first second."" 

Solution. From (10) and (11), 

(a) Vx = Vi cos 4>; t>y = ©i sin — 32.2 1 
Also, from (12), v = Vrja _ 64.4 1 Vi ain + 1036.8 «». 
From (16) and (16), oa- = ; Oy = - 82.2 ; o = - 32.2. 

(b) Substituting < = 1, uj = 100, = 30° in these results, we get 

t>x = 86.6 ft. per sec. ax = 0. 

Vf, = 17.8 ft. per sec. oy = - 82.2 ft. per (sec.)«. 

V = 88.4 ft. per sec. o = - 32.2 ft. per (sec.)«. 

u 17 ft 

(c) T = arc tan -^ = arc tan — ^ = 11°36'.9 = angle of direc- 
tion of motion with the horizontal. 




SIMPLE APPLICATIONS OF THE DERIVATIVE 107 

3. If a projectile be given an initial velocity of 200 ft. per sec. in a direction 
inclined 45° with the horizontal, find 

(a) the magnitude of the velocity and direction of motion at the end of the third 
and sixth seconds ; 

(b) the component velocities at the same instants. 
Conditions are the same as for Ex. 2. 

Ans, (a) When t = 3, t> = 148.3 ft per sec, t= 17^85', 
when t = 6, t> = 160.5 ft. per sec., t = 159® 53'; 
(b) when t = 8, r, = 141.4 ft. per sec., Vp = 44.8 ft. per sec. 
when i = 6, «« = 141.4 ft. per sec., ©^ = — 51.8 ft. per sec. 

4. The height (= s) in feet reached in t seconds by a body projected vertically 
upwards with a velocity of Vi ft. per sec. is given by the formula 

8 = tJi«-16.1t«. 

Find (a) velocity and acceleration at any instant; and, if ui = 300 ft per sec., 
find velocity and acceleration (b) at end of 2 seconds ; (c) at end of 16 seconds. 
Resistance of air is neglected. Ans. (a) « = ©i — 32.2 «, o = — 32.2 ; 

(b) V = 235.0 ft. per sec. upwards, 

a = 32.2 ft per (sec.)* downwards ; 

(c) V = 183 ft per sec. downwards, 

a = 32.2 ft. per (sec.)* downwards. 

6. A cannon ball is fired vertically upwards with a muzzle velocity of 644 ft 
per sec. Find (a) iU velocity at the end of 10 seconds ; (b) for how long it will 
continue to rise. Conditions same as for Ex. 4. 

Ans, (a) 322 ft. per sec. upwards ; 
(b) 20 seconds. 

6. A train left a station and in t hours was at a distance (space) of 

« = «» + 2«« + 3t 
miles from the starting point. Find its acceleration* (a) at the end of t hours; 
(b) at the end of 2 hours. Ans. (a) o = 6< + 4 ; 

(b) a = 10 miles per (hour)*. 

7. In t hours a train had reached a point at the distance of it* — 4t* + 10<» 
miles from the starting point (a) Find its velocity and acceleration, (b) When 
will the train stop to change the direction of its motion ? (c) Describe the motion 

during the first 10 hours. ^ ^ ^ ^ , .^ 

^ Ana. (a) t> = t« - 12t* + 32t, o = 3<* - 24< + 32 ; 

(b) at end of fourth and eighth hours ; 

(c) forward first 4 hours, backward the next 

4 hours, forward again after 8 hours. 

8. The space in feet described in t seconds by a point is expressed by the formula 

« = 48«-16t*. 

Find the velocity and acceleration at the end of li seconds. 

Ans. r = 0, o = - 32 ft. per (sec)*. 

• In this and th« following examples the magnitudes only of velocity and acceleration are 
required. 



108 DIFFERENTIAL CALCULUS 

9. Given a = 2 1 + 8 1> + 4 f* f t. ; find velocity and acceleration (a) at origin ; 
(b) at end of 6 seconds. Ana. (a) o = 2 ft. per sec., a = 6 f t. per (sec.)^ ; 

(b) c = 832 ft per sec., a = 126 ft per (sec.)«. 

10. Given s = - + bf^, where a and b are constants ; find velocity and accelera- 
tion at any instant Ana. v = h2M, a = \-2b. 

11. At the end of t seconds a body has a velocity of 3 ^ + 2 1 ft per sec. ; find 
its acceleration (a) in general ; (b) at the end of 4 seconds. 

^718. (a) a = 6 1 + 2 ft per (sec.)^ ; (b) a = 26 ft. per (sec.)>. 

12. The vertical component of velocity of a point at the end of t seconds is 

«y = 8«? - 2f + 6 ft per sec. 
Find the vertical component of acceleration (a) at any instant ; (b) at the end of 
2 seconds. Ana. (a) a, = 6t — 2 ; (b) 10 ft. per (sec.)^. 

13. If a point moves in a fixed path so that 

a = Vt, 
show that the acceleration is negative and proportional to the cube of the velocity. 

14. If the distance in feet described by a point in t seconds is given by the formula 

« = 101og , 

4 + t 

find velocity and acceleration (a) at the end of 1 second ; (b) at the end of 16 seconds. 

Ana. (a) c = — 2 ft per sec, o = J ft per (sec.)^; 

(b) r = — i ft per sec, o = ,^ f t. per (sec.)*. 

16. If the space described is given by 
show that the acceleration is always equal to the space passed over. 

16. Given s = a cos — : find acceleration. Ana. a = 

2 4 

17. If a point referred to rectangular coordinates moves so that 

X = a cos £ + &, and y = a sin f + c, 
show that its velocity has a constant magnitude. 

18. If the path of a moving point is the sine curve 

(x = a^, 

( y = 6 sin o^, 
show (a) that the x component of the velocity is constant ; (b) that the acceleration 
of the point at any instant is proportional to its distance from the axis of X. 

19. If a particle moves so that 

« = r», y = <«, 

(a) show that the path is the semicubical parabola y^ = x^; 

(b) find Vxy v^, V ; 

(c) find Ox, o^, a ; 

(d) when t = 2 sec, find v, a, position of point (coordinates), and direction of 
motion. 



CHAPTER Vin 
SUCCESSIVE DIFFERENTIATION 

87. Definition of successive derivatives. We have seen that the 
derivative of a function of x is in general also a function of a*. 
This new function may also be differentiable, in which case the 
derivative of the jir%t derivative is called the second derivative 
of the original function. Similarly the derivative of the second 
derivative is called the third derivative; and so on to the nth 
derivative. Thus, if 

dx 
dx\axj 

dx\jdx\dxj^ 

88. Notation. The symbols for the successive derivatives are 
usually abbreviated as follows : . 

dx\dxj do(^ 

dx\_dx\dx)^ dx\dx'J da? 



d_ /d'-'yN ^ ^ 

dxydx'-y dxr' 

If 1/ =f(T), the successive derivatives are also denoted by 

/'(^). /"(^), f"'{^), n^)^ ■■; r'H^)i 

f/^^)' £-^<^>' y^'^^ £^-^(^)' •••' ^•^<^>' 

109 



110 DIFFERENTIAL CALCULUS 

89. The lyth derivative. For certain functions a general expres- 
sion involving n may be found for the nth derivative. The usual 
plan is to find a number of the first successive derivatives, as many 
as may be necessary to discover their law of formation, and then 
by induction write down the nth derivative. 



Ex. 1. Given y = e^i find — . 

Solution. — = oc"*, 

dx 



— i = a'^e^. Ana. 



Ex. 2. Given y = logx ; find 



dx» 

d*y 
dx* 



Solutum. — = -i 

dz X 

dx^ x^ ' 



d»y_l-2 
dx' x« 

d^ _ 1 . 2 » 3 
dx*" X* ' 

• • • • 

d"i/ In — 1 

^ = (-1)— 1!=. Am. 

dx" x» 



Ex. 3. Given v = sin x ; find — - . 

dx» 

Solution. — = cos X = sin (X + - ) 

dx \ 2/ 



= — sinfx + — ) = C08( X + - ) = Bin(x H — -\ 
dz \ 2/ \ 2/ \ 2 / 



dxa 

»y d . / , 2t\ / , 2t\ . / , 3t\ 

= -sini X H I = COS! X H 1 = smi x H li 

c«dx \2/ \2/ \2/ 



-^ = ami X H ). Ana. 

dx« \ 2 / 



SUCCESSIVE DIFFERENTIATION 111 

90. Leibnitz's formttla for the nth derivative of a product. This 
formula expresses the nth derivative of the product of two variables 
in terms of the variables themselves and their successive derivatives. 

If u and V are functions of Xj we have, from Y, 

d , . du ' dv 
dx dx , dx 

Differentiating again with respect to Xy - 

cP , . cPiA du dv , du dv dFv 

— iuv) = V H 1 h u — 

d^? da? dx dx dx dx do? 

ePw , c%du dv ^ ePv 

da? ^ dxdx^ dj? 
Similarly 

d* , . d?u . dSi dv . rt d?u dv , g. du d?v du d'v , d?v 

— (wt;) = — V H 1- 2 h 2 1 h u — 

do? da? da? dx da? dx dx do? dx da? da? 

d?u . „ d?u dv „ du d?v d?v 

= — V '\' 6 h o h u — • 

da? da? dx dx da? da? 

However far this process may be continued, it will be seen that 
the numerical coefficients follow the same law as those of the 
Binomial Theorem, and the indices of the derivatives correspond 
to the exponents of the Binomial Theorem.* Reasoning then by 
mathematical induction from the wth to tlie (m + l)th derivative 
of the product, we can prove Leibnitz^ % Formula 

^ ' dx^^^^^ ■" dx^ ^^'^ dx^-^ dx "^ [2 €te~-« cte« "•■ ' " 

du d^-^v ', d'^i^ 
^ dxdx^'^^ dxn 

Ex- 1. Giyen y = e* log x ; find — ^ by Leibnitz^s Formula. 



SolutiOTi, Let 


tt = e*, 


and V = logx; 


then 




dv 1 
dx " X 




dx^ 


cPc 1 
dx2" X«* 



dhi dH 2 

— = c*, — = — . 

dx' dx* X? 

* Tb make thii oorrespondenoe complete, u and v are oonaidered as —- and •-— < 



112 DIFFERENTIAL CALCULUS 



Substitating in (17), we get 

8 



= ^(logx + --- + -) 



d"v 
Ex. 2. Given y = ««e«*; find — ^ by Leibnitz's Formula. 



Solution. Let 








tt = «», 


and c = e«« ; 


then 








d^ 


dx 
d^ 

d'o 
dx« 




d"u 
dx» 


d*^ 
dx"^ 


Substituting in 


(17), 


we 


get 








c^y 


— T 


^«« 


«e4- ^nnn- 


-Wfox 4- n^n _ 



dx" ' 

= a»-«c«*[x«a« + 2nax + n(n-l)]. 

91. Sttccessive differentiation of implicit functions. To illustrate 
the process we shall find -— ^ from the equation of the hyperbola 

6 V - ay = a*6». 
Differentiating with respect to a: as in § 75, p. 84, 

2 6»a:-2aVl^ = 0, or, 

M) ^ = ^ 

Differentiating again, remembering that y is a function of x^ 

^__ dx 



SUCCESSIVE DIFFERENTIATION 113 



dy 
Substituting for -^ its value from (-4.), 



epy 



^ \a'y) ^ V{h'2?-^a^f) 



da? a^y^ aV 

But from the given equation, 6 V — ay = (j?h\ 

da? a^y* 



EXAMPLES 

Differentiate the following. 

1. y = 4a!*-6x2 + 4x + 7. ^ = 12(2x-l). 

ox* 

2. f{x) = ^. /iT(x) = 



1 - X " ' ' (1 - X)* 

3. /(y) = j^. /^(y)=l6. . 

4. y = x« log X. 



d^y 6 



da:* X 
K _ c ^ — n(n-f l)c 

6. y = (x-3)^* + 4xe» + x. -4 = 4e'r(a; - 2)c* + x + 2] 

a. - --V (T^ 1 , - --, y 

2^ ' dx^ 2a^ ' a» 

8. /(x) = 0x9 + 6x + c. 

9. /(x) = log(x + l). 

10. /(x) = log(e* + e-*). 

11. r = sin o^. 

12. r = tan *. — - = 6 sec* — 4 sec* 0. 

d0« 

13. r = log sin 0. — = 2 cot cosec' 0. 

(i0' 

14. f(t) = e-'co8«. /iv(Q = -. 4e-'cos« = - 4/(t). 



dx2 


2a^ 


; - 


r'w = 


0. 




/*Ma^) = 


6 




(X + 1)* 




r'(x) = 


8 (c - e- 


x)8 


de* 


a* sin od = 


a*r. 



15. f{e) = V8ec2^. /" (^) = 3 [f(0)y - f(0). 



114 DIFFERENTIAL CALCULUS 



16. p = (g« + a2)arcUn^ ' - 



a dq* (a=» + q^* 

17. y = a*. 3^ = (log a)» a*. 



18. y = log(l+«). 



dx« 

d»v |n-l 
rii? = (-i)«-i ' , 

d*i/ / nT \ 

19. v=rco8ax. -~ = a»cosiaxH )• 

dx" \ 2 / 

20. y = x»-Uogx. -i^ = i=. 

OX'* X 

[nm a poeitive integer.] 

21. y = l=^. ^ = 2(-l)«— i? 

^ 1 + x 'dx» ^ ' (l + a)"+* 

2 
Hint. Reduce fraction to form - 1 + z before differentiating. 

22. If y = e=^8inx, prove that ~|-2-^ + 2y = 0. 

dx^ dx 

dhf dy 

23. Uy=z acoB (log x) + 6 sin (log x), prove that x^— ~H-x~ + y = (X 

dx^ dx 

Use Leibnitz^s Formula in the next four examples. 

24. y = x«a*. t^ = <»* (log a)""' [(« log a + n)J» - n]. 

dx" 

d*y 

26. y = e*ix. --^ = c*(x + n). 

dx*» 

•26. /(x) = c«sinx. /(")(x) = (V2)'«e«8in(x + — )• 

27. f{e) = cos o^ COB be. /C»>(^) = <?±^ cos [(a + 6) ^ + — 1 

, (a - 6)" r . , . _ , nr"! 
+ ^ COS [^(a - 6) ^ + _ J . 

28. Show that the formulas for acceleration, (14), (15), p. 105, may be written 

_d2« _ _d*x _ _d^ 
"""d£2 

29. y« = 4ax. 

30. Wx2 4. a2y3 = a2ft2. 

31. xa + y« = A 

32. y^ + y=:x*. 

dx« (l + 2y)* 



df'i' " d«2* 




d2y 4a* 




dx* y» 




d»y _ &* . d«y _ 


S¥x 


dxa~ a2y8' dx«~ 


aV 


d2y t^ 
dx* y« 




d»y 24 X 





SUCCESSIVE DIFFERENTIATION 115 



33. ax«H-2Axy + 6y« = l. 



38. €« + tt = e" + «. 



41. y« + «'-8axy = 0. 



(Py_ h^- ab 
dxa "" (Ax + by)* 



34. ya-2xy = a«. _=__; _ = -__. 

cP^ tana e - ten* 
d02 tan* $ 

d^e 2(6 + 8^ + 8^) 

36. tf = ten(0 + (^). 5^ = --^^ ^5 

cPc 4(tt + t)) 

37. log(u + r) = u-t,. _ = ___. 






39. a = l + «e.. _ = __^-. 

<J2« (2-a)c + 2« 

40. c" + rt-c = 0. 3^5 = * 



dp (C + 0» 

d^y _ 2 a»zy 
^■" (y«-ax)»' 



d»y a(m«-l) 

42. y.-2»«y + ««-a = 0. _ = ^_--^,. 

d?y — y 

43. y = 8in(x + y). Tl = 7; ; — i — ^15* 

9 \ "f dx» [1 — cos (X + y)]» 

d^_ y[(x-l)« + (y -!)»] . 

44. e»+. = xy. ^= ^^^-^ji 

46. ax« + 2Axy + 6y« + 2fifX4-2/y + c = 0. 



CHAPTER IX 



MAXIMA AND MINIMA* 



92. Increasing and decreasing functions. A function is said to be 
increaBing when it increases as the variable increases and decreases 
as the variable decreases. A function is said to be decreasing when 
it decreases as the variable increases and increases as the variable 
decreases. 

The graph of a function indicates plainly whether it is increas- 
ing or decreasing. For instance, consider the function a* whose 
graph (Fig. a) is the locus of the equation 

y = a*. a > 1 

As we move along the curve from left to right the curve is m- 
ing^ i.e. as x increases the function (= y) always increases. There- 
fore a* is an increasing function for all values of x. 



Y 




Ay 







X 




Fio. a 



Flo. 5 



On the other hand, consider the function (a — x)* whose graph 
(Fig. h) is the locus of the equation 

y = (« - ^f- 

* The proofs given in this chapter depend chiefly on geometric Intuition. The sul^ect of 
Maxima and Minima will he treated analytically in $ 119, p. 169. 

116 




MAXIMA AND MINIMA 117 

Now as we move along the curve from left to right the curve 
is falling^ i.e. as x increases the function (= y) always decreases. 
Hence (a — a;)' is a decreasing function for all values of x. 

That a function may be sometimes increasing and sometimes 
decreasing is shown by the graph (Fig. c) of 

y=2i»-9a? + 12a;-3. 

As we move along the curve from left to right the curve lises 
until we reach the point A^ then it falls from ^ to ^, and to the 
right of B it is always rising. Hence 

(a) from re = — oo to a; = 1 the function is 
increasing; 

(b) from x=zl to a: = 2 the function is de- 
creasing; 

(c) from x=2 to a? = -foo th^ function is 
increasing. 

The student should study the curve care- 
fully in order to note the behavior of the fio. « 
function when a? = 1 and a; = 2. Evidently A 
and B are turning points. At A the function ceases to increase 
and commences to decrease ; at B, the reverse is true. At A and 
B the tangent (or curve) is evidently parallel to the axis of X, and 
therefore the slope is zero. 

93. Tests for determining when a function is increasing and when 
decreasing. It is evident from Fig. c that at a point, as C, where 
a function ^. ^.^ 

is increasing^ the tangent in general makes an acute angle with the 
axis of X; hence 

slope = tan t = -~ =/' (x) = a positive number. 

dx 

Similarly at a point, as 2>, where a function is decreasing^ the tan- 
gent in general makes an obtuse angle with the axis of X\ therefore 

slope = tan t = 3^ =/' {x) = a negative number,* 

dx 

* ConTeraely, for any glren ralue of x, 

\ffXx)~+f then f{x) %9 incretMtng ; 

iff{z)='- , tken/(x) U decreasing . 
Wben/*(z)=0, we cannot decide without further InTestigation whether /(x) is increasing or 
decreasing. 



118 DIFFERENTIAL CALCULUS 

In order then that the function shall change from an increasmg 
to a decreasing function, or vice versa, it is a necessary and suffi- 
cient condition that the first derivatiye shall change sign. But this 
can only happen for a continuous derivative by passing through 
the value zero. Thus in Fig. c, p. 117, as we pass along the curve 
the derivative (= slope) changes sign at A and B where it has the 
value zero. In general then we have at turning points 

(18) ^=/r(x) = 0. 

The derivative is continuous in nearly all our important appli- 
cations, but it is interesting to note the case when the derivative 
(= slope) changes sign by passing through oo.* This would evi- 
dently happen at the points B^ E, G in Fig. d, p. 119, where the 
tangents (and curve) are perpendicular to the axis of X At such 
exceptional turning points 

or, what amounts to the same thing, 

1 



/(^) 



= 0. 



94. Maximum and minimum values of a function, f A maximum 
value of a function is one that is greater than any values imme- 
diately preceding or following. 

A minimum value of a function is one that is less than any 
values immediately preceding or following. 

For example, in Fig. c, p. 117, it is clear that the function has 
a maximum value MA (= y = 2)^when a? = 1, and a minimum value 
NB{=i/= 1) when x=2. 

The student should observe that a maximum value is not neces- 
sarily the greatest possible value of a function nor a minimum 
value the least. For, in Fig. c it is seen that the function (= y) has 
values to the right of B that are greater than the maximum MA^ 
and values to the left of A that are less than the minimum NB. 

* By this is meant that its reciprocal passes tlirouKli the value zero. 

t The student should not forget that in general the definitions and proofs giren in this book 
apply only at points where the function is continuous. 



MAXIMA AND MINIMA 



119 



A function may have several maximum and minimum values. 
Suppose that the following figure represents the graph of a 
function f{x). 

At B^ Z>, G, /, K the function is a maximum, and at C, E^ jfiT, J 
a minimum. That some particular minimum value of a function 
may be greater than some particular maximum value is shown in 




FlO. d 



the figure, the minimum values at C and H being greater than the 
maximum value at K. 

At the ordinary turning points C, 2>, H^ /, Ji K the tangent 
(or curve) is parallel to OX; therefore 

«Zoj9e = ^=/'(2r) = 0. 

At the exceptional turning points B^ E^ G the tangent (or curve) 
is perpendicular to OX, giving 



%lope 



= ^ = 



dx 



f{x) = oo. 



One of these two conditions is then necessary in order that 
the function shall have a maximum or a minimum value. But 
such a condition is not sufficient ; for, at F the slope is zero and at 
A it is infinite, and yet the function has neither a maximum nor a 
minimum value at either point. It is necessary for us to know, in 
addition, how the function behaves in the neighborhood of each 
point. Thus at the points of maximum value^ B, Z>, Gj /, K, the 
function changes from an increasing to a decreasing function^ and 
at the points of minimum value^ C, E^ if, Ji the function changes 



120 DIFFERENTIAL CALCULUS 

frtym a decreasing to an increasing function. It therefore follows 
from § 93 that at maTimum points 

slope = 3^ =f'(x) must change from + to — » 
dx 

and at minimum points 

slope =-^ =f\x) must change from — to 4 

when we move along the curve from left to right. 

At such points as A and i^ where the slope is zero or infinite, 
but which are neither maximum nor minimum points^ 

slope = -~- =/'(2:) does not change sign, 
dx 

We may then state the conditions in general for maximum and 
minimum values oif{x) for certain values of the variable as follows : 

(19) f(x) is a maximum if /'(x) = O, and f(Qc) chansres from + 
to — . 

(20) f(x) is a minimum if /'(x) = O, and f^(x) chansres from — 
to +• 

The values of the variable at the turning points of a function 
are called critical values; thus x=\ and x = 2 are the critical 
values of the variable for the function whose graph is shown in 
Fig. (7, p. 117. The critical values at turning points where the 
tangent is parallel to OX are evidently found by placing the first 
derivative equal to zero and solving for real values of x^ just as 
under §77, p. 86.* 

To determine the sign of the first derivative at points near a 
particular turning point, substitute in it, first a value of the variable 
just a little less than the corresponding critical value, and then one 
a little greater.^ If the first gives + (as at X, Fig. rf, p. 119) and 
the second — (as at M)^ then the function (= y) has a maximum 
value in that interval (as at /). 

* Similarly if we wish to examine a function at exceptional turning points where the tangent 
is perpendicular to OX, we set the reciprocal of the first deriyatire equal to aero and solve to 
find critical values. 

t In this connection the term " little less," or " trifle less,** means any value between tlio 
next smaller root (critical value) and the one under consideration ; and the term " little greater,*' 
or " trifie greater," means any value between the root under consideration and the next larger onck 



MAXIMA AND MINIMA 121 

If the first gives — (as at P) and the second + (as at iV), then 
the function (= y) has a minimum value in that interval (as at C). 

If the sign is the same in both cases (as at Q and JS), then the 
function (= y) has neither a maximum nor a minimum value in 
that interval (as at F).* 

We shall now summarize our results into a compact working rule. 

95. First method for examining a function for maximum and 
minimum values. Working rule. 

First step. Find the first derivative of the function. 

Second step. Set the first derivative equal to zero f and solve the 
resulting equation for real roots in order to find the critical values of 
the variable. 

Third step. Write the derivative in factor form; if it is algebraic^ 
write it in linear factor form. 

Fourth step. Considering one critical value at a iimcy test the 
first derivative^ first for a value a trifle less and then for a value a 
trifle greater than the critical value. If the sign of the derivative is 
first -f and then — , the function has a maximum value for thai partic- 
ular critical value of the variable; but if the reverse is true^ then 
it has a minimum value. If the sign does not change^ the function 
has neither. 

Ex. 1. Examine the function (x — 1)^(2 + 1)' for maximnm and minimum 
values. 

SoLxdvoTL f(z) = (X - 1)« (X + 1)». 

Firststep. /'(x)=2(x-lXa; + l)H3(x~l)2(x + l)2 J 

= (X - l)(x + l)«(6x - 1). ..i^X 

Second step. (x-l)(x+ 1)«(6» -1)=0, ^ 

X = 1, — 1} i, whicli are critical yalues. 

Third step. /'(x) = 5(x - 1) (x + l)«(x - i). 

Fourth step. Examine first for critical value x = 1 (C in figure). 

Whenx<l,r(x) = 5(-) (+)«(+) = -. 

When X > 1, r{x) = 5(+)(+)«(+) = +. 

Therefore, when x = 1 the function has a minimum value /(I) = (= ordinate 
of C). 

* A similar disctusion will evidently hold for the exceptional turning points £, E, and A 
respectlrely. 

t When the first derlTatire becomes infinite for a certain ralue of the independent variable, 
then the function should be examined for such a critical value of the variable, for it may give 
maximum or minimum values, as at B, E, or A (Fig. d, p. 119). See footnote on preceding page. 



122 DIFFERENTIAL CALCULUS 

Examine now for critical value x = | (B in figure). 

When x<i, r(x) = 6(-) (+)2(-) = +• 
Whenx>J,/'(x) = 5(-)(+)2(+)=-. 

Therefore, when x = J the function has a maximum value /(}) = 1.11 + (= ordi- 
nate of B). 

Examine lastly for critical value x = — 1 (^ in figure). 

Whenx<-l,/'(x) = 6(-)(-)5»(-) = +. 
When X > - 1, r(x) = 5(-)(+)2(-) = +. 

Therefore, when x = — 1 the function has neither a maximum nor a minimum 
value. 

Ex. 2. Examine sin x (1 + cos x) for maximum and minimum values. 

Solvtion. f{x) = sin x (1 + cos x). 

First step. f'(x) = -sin2x-|-(l + cosx)co8X=2coa^x + coBX- 1. 

Second atep. 2 cos^ x + cosx — 1 = 0. 

Solving the quadratic, cos x = J or — 1 ; 

hence the critical values are x = i: - or t. 

3 

Third step. f\x) = 2 (cos x - i) (cos x + 1). 

Fourth, step. Examine first for critical value x = — 

3 

When x<|, /'(x) = 2 (+)(+)=+. 

When x>^, /'(x) = 2 (-)(+) = -. 

Therefore, when x = ^ the function has a maximum value /(-) = - VS. 

3 \3/ 4 

Examine now for critical value x = • 

3 

Whenx<-^, /'(x) = 2 (-)(+) = -. 

When x> - ^, f\x) = 2(+)(+) = +. 

Therefore, when x = - - the function has a minimum value /(--) = — Vs. 

3 V 3/ 4 

Examine now for critical value x = ir. 

When x<x, /'(x) = 2 (-)(+) =-. 
Whenx>x, /'(x) = 2 (-)(+)= -. 

Therefore, when x = «- the function has neither a maximum nor a minimum 
value. 




MAXIMA AND MINIMA 123 

Since the cosine is a periodic function, the critical yalues are really 

X = 2 nir ± - and nr, 

where n is any integer. Therefore the function has an infinite number of maxima 
all equal to ] V3, and an infinite number of minima all equal to — } V3. 

£x. 3. Examine the function a - 6 (x — c)' for maxima and minima. 
Solution. f(x) = a — 6(x — c)K 

3(x-c)* 

Since x = c is a critical value for which /'(x) = 00, but 
for which /(x) is not infinite, let us test the function for 
maximum and minimum values when x = c. 

When x<c, /'(x) = +. 

When X > c, /'(x) = - . 

Hence, when x = c = OM the function has a maximum value /(c) = a = MP, 

96. Second method for ezamining a function for maximum and 
minimum values. From (19), p. 120, it is dear that in the vicinity 
of a maximum value of /(a:), in passing along the graph from left 

f\x) changes from -{- to to ^,* 

Hence /'(x) is a decreasing function, and by § 93 we know that its 
derivative, i.e. the second derivative of the function itself [=/"(a:)], 
is negative or zero. 

Similarly we have, from (20), p. 120, that in the vicinity of a 
minimum value of f{x) 

/'(a:) changes from — to to +. 

Hence f\x) is an increasing function, and by § 93 it follows that 

f'\x) is positive or zero. 

The student should observe that /"(a:) is 

positive not only at minimum points (as at A) 

but also at points such as P. For, as a point 

^ passes through P in moving from left to right, 

slope = tan t = -~ =f(x) is an increasing function. 

dx 

At such a point the curve is said to be concave upwards. 

*f{x) is aasumed to be contintiouB, and/^(x) to exist. 




124 



DIFFERENTIAL CALCULUS 




Similarly /"(x) is negative not only at 
maximnm points (as at B) but also at points 
such as Q. For, as a point passes through Q, 



glope = tan t = -~ =/' (x) in a deerea9inff function, 

dx 

At such a point the curve is said to be concave downwards,* 

We may then state the sufficient conditions for maximum and 
minimum values of f{x) for certain values of the variable as 
follows : 

(21) /(x) is a nuudmum if ff(x) = O and ff^(x) = a nefirative 
number. 

(22) f(x) is a minimum if f^(x) = O and /^^(x) = a positive 
number. 

Following is the corresponding working rule. 

First step. Find the first derivative of the function. 

Second step. Set the first derivative equal to zero and solve the 
resulting equation for real roots in order to find the critical values of 
the variable. 

Third step. Find the second derivative. 

Fourth step. Substitute each critical value for the variable in the 
second derivative. If the result is negative^ then the function is a 
maximum for that critical value ; if the result is positive^ the function 
is a minimum, jf 

Ex. 1. Examine x* — 3 x^ — z + 6 for maxima and minima. 

SolutUm. /(x) = x«-3x«-9x + 6. 

First step. f(x) = 3x« - 6x - 9. 

Second step. 8x* - 6x - 9 = 0; 

* hence the critical values are x = — 1 and 3. 

Third step. /"(x) = 6x-6. 

Fourth step. /''(-I) = -12. .-./(-I) = 10 = (ordinate of 
A) = maximum value. 

/" (3) = + 12. .•./(3) = - 22 (ordinate of B) 
= minimum value. 

* At a point where the cuxre is concave upwards tre sometimes say that the carve has & positive 
bending^ and where it is concave dotontvarda a negative bending. 

t When/"(x)sO, or does not exist, the above process fails, although there may even then be 
a maximam or a minimum ; in that case the first method given in the last section still holds, 
being fundamental. Usually this second method does apply, and when the process of flniUnq 
the second derivative is not too long or tedious, it is generally the shortest method. 





MAXIMA AND MINIMA 126 

Ex. 2. Examine sin' x cosx for maximum and wtinimnm valaes. 

SolutioTL f{z) = sin^ z cos x. 

Fint step. f(z) = 2 sin z co8» x — sin' x. 

Second step, 2 sin x cos^ x — sin" x = ; "^ 

hence the critical values are x = nr ^ 

and X = nr i: arc tan VS = nx ± o. 

Third step, f* (x) = cos x (2 cos^ x - 7 sin« x). 

Fouxik step, /" (0) = +. .'./(O) = = minimum value at 0. 

/"■ (x) = — . .'. /(ir) = = maximum value at C. 

/" (o) = — . .-. /(o) = maximum value at A, 

/" (x — a) = + . .*. /(x — o) = minimum value at B, etc. 

The work of finding maximum and minimum values may frequently be simpli- 
fied by the aid of the following principles which follow at once from our diroussion 
of the subject. 

(1) The maximum and minimum values of a continuous function must occur 
aUernatdy, 

(2) When c is a positive constant, cf(z) is a maximum or a minimum for such 
values of X, and such only, as make f{z) a maximum or a minimum. 

Hence, in determining the critical values of x and testing for maxima and minima, 
any constant factor may be omitted. 

When c is negative, cf{z) is a maximum v)henf(z) is a minimum, and conversely, 

(3) Ifc is a constant, -, . , , ^, . 
^ ' f(x) and c +/(x) 

have maximum and minimum values for the same values ofz* 

Hence a constant term may be omitted when finding critical values of x and 
testing. 



Examine the following functions for maximum and minimum values. 

1. 8x«-9x«-27x + 30. Arvi, x = - 1, gives max. =45; 

X = 3, gives min. = — 61. 

2. 2x«-21xa + 36x-20. AnA, x = 1, gives max. = - 8 ; 

X = 6, gives min. = — 128. 

x" 
3. 2x* + 3x + l. • An», x = 1, gives max. ={; 



3 



X = 8, gives min. = 1. 



4. 2x»-16x« + 86x + 10. An», x = 2, gives max. = 38 ; 

X = 3, gives min. = 37. 

5. x* — 9x* + 15x — 3. An», x = 1, gives max. = 4 ; 

X = 5, gives min. = — 28. 

8. x* — 3x^ + 6x + 10. Anjs, No max. or min. 

7. X* ~ 5x* + fix* + 1. An», x = 1, gives max. = 2 ; 

X = 3, gives min. = — 26 ; 
X = 0, gives neither. 



126 



DIFFERENTIAL CALCULUS 



8. 3z»~ 126x8 + 2160X. 



9. (x-3)2(x-2). 
10. (x-l)8(x-2)2. 



11. (z - 4)6 (X + 2)*. 



12. (x-2)6(2x+l)*. 



13. (x + l)'(x-6)2. 

14. (2x-a)*(x-a)l 



15. «(x-l)«(x + l)«. 

16. X (a + x)« (a - x)«. 

17. 6 + c(x -a)'. 

18. a-b(x-c)K 

19. ^^-7x + 6 

20. 
21. 
22. 

23. 



X 


- 10 • 


(a- 

a - 


-X)8 

2x 


1- 


z + x« 


1 + 


x-x* 


X2- 


■3x + 2 


x2 + 3x + 2 


{X- 


-a)(6-x) 




X2 


^2. 


6a 



X a — X 



^na. X = — 4 and 3, give max. ; 
X = — 8 and 4, give mln. 

Am, X = }, gives max. = ^ ; 
X = 3, gives min. = 0. 

Aivi, X = j, gives max. = .03456 ; 
X = 2, gives min. = ; 
X = 1, gives neither. 

Am. X = — 2, gives max. ; 
X = f , gives min. ; 
X = 4, gives neither. 

Ai\», X = — ^, gives max. ; 
X = II, gives min. ; 
X = 2, gives neither. 

Am, X = If gives max. ; 

X = — 1 and 6, give min. 

2a 
Am, X = — f gives max. ; 

X = a, gives min. ; 

X = -» gives neither. 
2 

Ana. X = ^, gives max. ; 

X = 1 and — I, give min. 

An8. X = — a and - , give max. ; 

o 

X = — » gives mm. 
2 

Ana. X = a, gives min. = 6. 
Am, No max. or min. 

Am. X = 4, gives max. ; 
X = 16, gives min. 

Am. X = - f gives min. 
4 

Ana. X = }, gives min. 

Ana. X = V2, gives min. = 12 V2 — 17 ; 

x=— V2, givesmax. = — 12v^-17. 

. 2ab (a-6)a 

Am. X = ri gives max. = - - - 



a + h 



4a6 



Ana. x = 



a'^ 



« = 



a -- 6 



a + 6 



t gives min. ; 



t gives max. 



26. 


X 

logx 


26. 


X 



MAXIMA AND MINIMA 127 

Ans. X = e, gives min. 

Ana. MiD. when x lies between ^f and |}. 

27. ae** + be-**. Ana. Min. = 2 VoS. 

28. X'. Ana. x = -y gives min. 

e 

29. x^. Ana. x = e, gives max. 

30. COB X + sin X. Ana. x = - » gives max. = V2 ; 

4 

X = — - , gives min. = — V2. 
4 

31. sin 2 X — X. ^n«. x = - 1 gives max. ; 

X = ~ — I gives min. 
6 

32. X + tan x. ^n«. No max. or min. 

T 3 /- 

33. sin^x cos x. ^tu. x = - , gives max. = — v3 : 

3 ^ 16 

r . . 3 /r 

X = — -- , gives mm. = VS. 

3 * 16 

34. xcosx. ^715. X = cot X, gives max. 

36. sin X + cos 2 x. Ana. x = arc sin |, gives max. ; 

X = - , gives min. 

36. 2 tan X — tan^x. Ana. x = - ) gives max. 

4 

-_ sinx A ^ • 

37. . j^ns. x = —i gives max. 

1 + tanx 4 

X 

38. Ana. x = cos x, gives max. 

1 + xtanx 

39. The range OA of a projectile in a vacuum is given by the formula 

_, ri2sin20 
Ji = ; 

9 
where Vi = initial velocity, g = acceleration due to grav- 
ity, ^ = angle of projection with the horizontal. Find 
the angle of projection which gives the greatest range o 
for a given initial velocity. Ana. = 46®. 

40. The total time of flight of the projectile in the last problem is given by the 

formula _ 2tJi8m0 

J^ = . 

9 
At what angle should it be projected iu order to make the time of flight a 

maximum ? Ana. 4> = 90®. 




128 DIFFERENTIAL CALCULUS 

41. The time it takes a ball to roll down an inclined plane AB is giyen by the 
a formula 




^^^ r 



if08m2^ 



or sin 20 

Neglecting friction, etc., what must be the yalue of to make the 
r^ quickest descent ? Aia. ^ = 46°. 

42. When the resistance of air is taken into account, the inclination of a pen- 
dulum to the vertical may be given by the formula 

$ = ac-** cos (tU + e). 

Show that the greatest elongations occur at equal intervals - of time. 

n 

43. It is required to measure a certain unknown magnitude x with precision. 

Suppose that n equally careful observations of the magnitude are made, giving the 

results 

ai, Os} da* • • • » flu* 

The errors of these observations are evidently 

X — fli, X — Osi X — Os, • • • , X — Oa, 

some of which are positive and some negative. 

It has been agreed that the most probable value of x is such that it renders the 
sum of the squares of the errors, namely 

(X - ai)« + (X - a,)«+ (X - aa)a + . . . + (x - a,)«, 

a minimum. Show that this gives the arithmetical mean of the observations as the 
most probable value of x. 

44. The bending moment at B of a beam of length I, uniformly loaded, is given 
by the formula 3f=i«rfx-lwx?, 

where w = load per unit length. Show that the maximum 
bending moment is at the center of the beam. 

45. If the total waste per mile in an electric conductor is 

r 

where c = current in amperes, r = resistance in ohms per mile, and f = a constant 
depending on the interest on the investment and the depreciation of the plant, what 
is the relation between c, r, and t when the waste is a minimum ? Ans. cr = t, 

46. A submarine telegraph cable consists of a core of copper wires with a cover- 
ing made of nonconducting material. If x denote the ratio of the radius of the core 
to the thickness of the covering, it is known that the speed of signaling varies as 

xMog-. 

X 

Show that the greatest speed is attained when x = — • 




MAXIMA AND MINIMA 129 

47. Afisaming that the power given out by a voltaic cell Is given by the formula 

where E = constant electro-motive force, r = constant internal resistance, B = exter- 
nal resistance, prove that P is a maximum when r = R, 

48. When a battery of mn cells is joined up so that m rows of n cells, connected 
in series, are joined in parallel, the current is given by the formula 

_ fiinE ^ 

{j =r — — , 

mR-^-wr 

where E = electro-motive force of each cell, r = internal, and S = external resist- 
ance of each cell. Show that the current is a maximum when Rm = m, that is, 
the total internal resistance equals the total external resistance. 

49. The force exerted by a circular electric current of radius a on a small magnet 
whose axis coincides with the axis of the circle varies as 

X 

where x = distance of magnet from plane of circle. Prove that the force is a maxi- 
mum when x = -• 

2 

60. We have two sources of heat at A and B with intensities a and h respectively. 
The total intensity of heat at a distance of x from A is given by the formula 

1 = -^+ * 



Show that the temperature at P will be the lowest when -4- F S 

d-x _Vb 

th^t is, the distances BP and AP have the same ratio as the cube roots of the 
corresponding heat intensities. The distance of P from A is 

aki 

x= 

a* + 6* 

97. General directions for solving problems involving maxima and 
minima. In our work thus far the function has been given whose 
maximum and minimum values were required. This is by no means 
always the case ; in fact, we are generally obliged to construct the 
function ourselves from the conditions given in the problem, and 
then test it as we have been doing for maximum and minimum 
values. 

No rule applicable in all cases can be given for constructing the 
function, but in a large number of problems we may be guided by 
the following 



130 DIFFERENTIAL CALCULUS 

General Directions. 

(a) Express the function whose maodmum or minimum i% involved 
in the problem, 

(6) If the resulting expression contains more than on^ variable^ the 
conditions of the problem will furnish enough relations between the 
variables so that all may be expressed in terms of a single one, 

{c) To the resulting functioii of a single variable apply one of our 
two rules for finding! maximum and minimum values, 

(d) In practical problems it is usually easy to tell which critical 
value will give a maximum and which a minimum value^ so it is not 
always necessary to apply the fourth step of our rules. 

m 

PROBLEMS 

1. It is desired to make an open-top box of greatest possible volume from a 
square piece of tin wbpse side is a by cutting equal squares out of the comers and 

then folding up the tin to form the sides. What should be 
the length of a side of the squares cut out ? 

SolvJtion. Let x = side of small square = depth of box ; 

then a — 2 X = side of square forming bottom of 

I) box, and volume is 

F = (a-2x)«x; 

^ which is the function to be made a maximum by varying x. 
Applying rule, 

dV 
First step. "T" = (« - 2x)2 - 4x(a - 2x) = a^ - 8ax + 12x«. 

ax 

Second step. Solving a^ — 8ax + 12x^ = gives critical values x = - and -• 

d 
It is evident from the figure that x = - must give a minimum, for then all the tin 

would be cut away, leaving no material out of which to make a box. By the usual 

a 2 a? 

test, X = - is found to give a maximum volume — • Hence the side of the square 
6 27 

to be cut out is one sixth of the side of the given square. 

2. Assuming that the strength of a beam with rectangular cross section varies 
directly as the breadth and as the square of the depth, what are the dimensions of 
the strongest beam that can be sawed out of a round log whose 
diameter is d ? 

Sohdion, If x = breadth and y = depth, then the beam will 
have maximum strength when the function xy^ is a maximum. 
From the figure, y^ = d^ — x^ ; hence we should test the function 

/(x) = X ((P - x*). 




MAXIMA AND MINIMA 



131 



First step. 



f(x) = - 2x3 + (P - x« = d» - 8»a 





a 
Second step, (P— 3 x2= 0. .*. x = -—= = critical value which gives a maximum. 

V3 
Therefore, M the beam is cat so that 

depth = Vf of the diameter of log, 

and breadth = Vi of the diameter of log, 

the beam will have maximum strength. 

3. What is the width of the rectangle of maximum 
area that can be inscribed in a given segment OAA* of a 
parabola ? 

Hint. If OC-A, BC'^h-x and PP'=»2y; therefore the area 

of rectangle PDJyP' is 

2{h-x)y. 

But since P lies on the parabola y* = 2px, the function to be tested is 

2ih-x)y/2px. 

Ans, Width = 1^ 

4. Find the altitude of the cone of maximum volume that 
can be inscribed in a sphere of radius r. 

Bint. Volume of cone- iira:»y. But x«= BCx CZ)« y (2r-y) ; there- 
fore the function to be tested is 

/(y)-^y'(2r-y). 

Ans, Altitude of cone = } r. 

6. Find the altitude of the cylinder of maximum volume 

that can be inscribed in a given right cone. 

Bint. Let AC^ r and BC= h. Volume of cylinder « nafly. 
But from similar triangles ABC and DBO 

K ^ r{h-y) 

r:x::h:h~y, .*. x= — - — * 

A 

Hence the function to be tested is 

/(y)=y(*-y)*. Ans. Altitude = i A. 

6. Divide a into two parts such that their product is a 

°>a^^^™- Ans. Each part = ?. 

2 

7. Divide 10 into two parts such that the sum of their squares is a minimum. 

Ans, Each part = 5. 

8. Divide 10 into two such parts that the sum of the double of one and square 
of the other may be a minimum. -^'M. 9 and 1. 

9. Find the number that exceeds its square by the greatest possible quantity. 

Ans. |. 

10. What number added to its reciprocal gives the least possible sum ? Ans. 1. 

11. Assuming that the stiffness of a beam of rectangular cross section varies 

directly as the breadth and the cube of the depth, what must be the breadth of the 

stiffest beam that can be cut from ft log 16 Uiches in diameter ? 

Ans. Breadth = 8 inches. 




132 DIFFERENTIAL CALCULUS 

12. A torpedo boat is anchored 9 miles from the nearest point of a beach, and 
it is desired to send a messenger in the shortest possible time to a military camp 
situated 15 miles from that pohit along the shore. If he can walk 5 miles an hour 
but row only 4 miles an hour, required the place he must land. 

An8» 3 miles from the camp. 

13. For a certain specified sum a man takes the contract to build a rectangular 
water tank lined with lead. It has a square base and open top, and holds 108 cubic 
yards. What shall be its dimensions in order to require the least possible quantity 
of lead ? AuB. Altitude = 3 yds., side = 6 yds., that is, 

length of side = twice the altitude. 

14. A gasholder is a cylindrical vessel closed at the top and open at the bottom, 
where it sinks into the water. What should be its proportions for a given volume 
to require the least material (this would also give least weight) ? 

Ana. Diameter = double the height. 

16. What should be the dimensions and weight of a gasholder of 8,000,000 cubic 
feet capacity built in the most economical manner out of sheet iron ^ of an inch 
thick and weighing 2^ lbs. per sq. ft.? 

Ana. Height = 137 ft., diam. = 273 ft, weight = 200 tons. 

16. What are the most economical proportions for a cylindrical steam boiler ? 

Ana. Diameter = length. 

17. A paper-box manufacturer has in stock a quantity of strawboard 30 inches 
by 14 inches. Out of this material he wishes to make open-top boxes by cutting 
equal squares out of each comer and then folding up to form the sides. Find the 
side of the square that should be cut out in order to give the boxes maximum 
volume. Ana. 3 inches. 

18. A roofer wishes to make an open gutter of maximum 
capacity whose bottom and sides are each 4 inches wide and 
whose sides have the same slope. What should be the width 
across the top ? Am. 8 inches. 



19. Assuming that the energy expended in driving a steamboat through the 
water varies as the cube of her velocity, find her most economical rate per hour 
when steaming against a current running c miles per hour. 

Hint, Let v - most eoonomlcftl speed ; 

then av* « energy expended each hour, a being a constant depending npon the psrticalar 

oonditionB, 

and V - e » actual distance advanced per hour. 

at* 
Hence is the energy expended per mile of distance advanced, and it Is therefore the 

v-c 

function whose minimum is wanted. Ana. v = { c. 

20. Prove that a conical tent of a given capacity will require the least amount of 
canvas when the height is V2 times the radius of the base. Show that when the 
canvas is laid out flat it will be a circle with a sector of about 66^ cut out A bell 
tent 10 ft. high should then have a base of diameter 14 ft. and would require 272 
sq. ft. of canvas. 



MAXIMA AND MINIMA 133 

21. Find the right triangle of maximum area that can be constructed on a line 

of length h as hypotenuse. . h , _., . ^ .». , 

° •' *^ Ana. —rz = length of both legs. 

22. Show that a square is the rectangle of maximum area that can be inscribed 
in a given circle. Also show that the square has the maximum perimeter. 

23. What is the isosceles triangle of maximum area that can be inscribed in a 
given circle ? Ans. An equilateral triangle. 

24. Find the altitude of the maximum rectangle that can be inscribed in a right 

triangle with base h and altitude fu .i « i... ^ ^ 

** Ans. Altitude = -• 

2 

25. Find the dimensions of the rectangle of maximum area that can be inscribed 
in the ellipse li^ + aV = a^ft^. jin^, a Vi and 6 V2 ; area = 2a6. 

26. Find the altitude of the right cylinder of maximum volume that can be 

inscribed in a sphere of radius r. , . ,^,^ j « 1. j 2 r 

^ Ans. Altitude of cylinder = —-1 . 

Vs 

27. Find the altitude of the right cylinder of maximum convex (curved) surface 
that can be inscribed hi a given sphere. ^^3. Altitude of cylinder = r V2. 

28. Find the altitude of the right cylinder inscribed in a given sphere when its 

entire surface is a maximum. . ..^.^ , /» 2 \i 

Ana. Altitude = 12 — j r. 

29. Find the altitude of the right cone inscribed in a sphere when its entire 
surface is a maximum. ^^ Altitude = (23 -VlT)^. 

30. Find the altitude of the right cone of minimum volume circumscribed about 
a given sphere. Ana, Altitude = 4r, and volume = 2 x vol. of sphere. 

31. A right cone of maximum volume is inscribed in a given right cone, the vertex 
of the inside cone being at the center of the base of the given cone. Show that the 
altitude of the inside cone is one third the altitude of the given cone. 

32. Through a point (a, b) referred to rectangular axes a straight line is to be 
drawn, forming with the axes a triangle of least area. Show that its intercepts on 
the axes are 2 a and 2 b, 

33. Through the point (a, b) a line is drawn such that the part intercepted 
between the axes is a minimum. Prove that its length is (a' + 5' )'. 

34. Given a point on the axis of the parabola ^ = 2|»: at a distance a from the 
vertex ; find the abscissa of the point of the curve nearest to it. Ans. x = a — p. 

35. What is the length of the shortest line that can be drawn tangent to the 
ellipse lAfl + oflffi = a^ and meeting the coordinate axes ? Ana. a + 6. 

36. Find the altitude of the least isosceles triangle that can be circumscribed 
about an ellipse, the base being parallel to the major axis. 

Ana. Altitude = 3 times semi-minor axis. 



134 DIFFERENTIAL CALCULUS 

37. A Norman window consists of a rectangle surmounted by a semicircle. 
Given the perimeter; required the height and breadth of the window when the 
quantity of light admitted is a maximum. 

Ana. Radius of circle = height of rectangle. 

38. A tapestry 7 feet in height is hung on a wall so that its lower edge is 9 feet 
above an observer's eye. At what distance from the wall should he stand in order 
to obtain the most favorable view ? 

Hint. The vertical angle Bubtended by the tapeetry in the eye of the obeerver must he at a 
maximum. An8. 12 feet 

39. The regulations of the British Parcels Post require that the sum of the length 
and girth of a parcel shall not exceed 6 feet. Show that 

(a) The greatest sphere allowed is about 17} inches in diameter and has a volume 
of about 1^ cubic feet. 

(b) The greatest cube has an edge 14} inches in length and a volume of nearly 
1| cubic feet. 

(c) The greatest rectangular box is 1 ft. by 1 ft. by 2 ft , containing 2 cubic feet 

(d) The greatest parcel of any shape is a cylinder 2 ft. long and 4 ft. cii'bum- 
ference, and contains over 2^ cubic feet 

40. What are the most economical proportions of a tin can which shall have a 

given capacity, making allowance for waste ? 

r J ] Bint. There is no waste in cutting out tin for the side of the can, 

^A^^A^.^^^,^^. but for top and bottom a hexagon of tin circumscribing the circular 
I 1^ I I P**®*" required is used up. 

^o^^^*^'^^^*^^ Ana. Height = x diameter of base. 

r 

Note 1. Ex. 16 shows that if no allowance is made for waste, then height 
= diameter. 

Note 2. We know that the shape of a bee cell is hexagonal, giving a certain 
capacity for honey with the greatest possible economy of wax. 

41. An open cylindrical trough is constructed by bending a given sheet of tin 

of breadth 2 a. Find the radius of the cylinder of which the trough forms a part 

when the capacity of the trough is a maximum. ^ _ , 2 a 

Ana. Rad. = — 

42. A weight IT is to be raised by means of a lever with the force F at one 
end and the point of support at the other. If the weight is suspended from a point 
at a distance a from the point of support, and the weight of the 

beam is w pounds per linear foot, what should be the length of 
the lever in order that the force required to lift it shall be a 

minimum? SlTJr, , 

Ana. X = V feet 

^ w 




43. A rectangular stockade is to be built which must have a certain area. If a 
stone wall already constructed is available for one of the sides, find the dimensions 
which would make the cost of construction the least 

Ana. Side parallel to wall = twice the length of each end. 



M.VXT MA AND MINIMA 135 

44. An electric arc light ta to be placed directly over the center of a ciicular 
plot o[ grass 100 feet [a diameter. Aasuming that the iDtenait; of light variea 
directly as tbe sine of the angle under which it strikes an illwninated surface and 
inversely as tbe square of its distance from the surface, how high should the liglit 
be hung in order that tbe best possible Ugbt shall fall on a walk along the circum- 
ference of the plot ? J 60 , . 

*^ Am. — T^feet 

v5 

45. Tbe lower comer of a leaf, whose widUi is a, is folded over 
Eo as just to reach the inner edge of page, (a) Find the width of 
the part folded over when the length of the crease is a mioimiun. 
(b) Find width when the area folded over la a mEnimuin. 

Ar^. W?.i(b)?a. 




CHAPTER X 

POIKTS OF nVFIECTION 

98. Definition. Potn^^c^f tn^6(?f ton separate arcs concave upwards 
from arcs concave downwards.* Thus, if a curve y =f{x) changes 

(as at B) from concave upwards (as at A) to 
concave downwards (as at C), or the reverse, 
then such a point as ^ is called a point of 
inflection. 

From the discussion of § 96 it follows at 
once that at A^ f\^) = +> and at C, f{x) = — . 
In order to change sign it must pass through the value zero;f 
hence we have 

(23) At points of in laection, ff(x)— O. 

Solving the equation resulting from (23) gives the abscissas of 
the points of inflection. To determine the direction of curving in 
the vicinity of a point of inflection, test f"{x) for values of a;, first 
a trifle less and then a trifle greater than the abscissa at that point. 

\if\x) changes sign, we have a point of inflection, and the signs 
obtained determine if the curve is concave upwards or concave 
downwards in the neighborhood of each point. 

The student should observe that near a point where the curve 
is concave upwards (as at A) the curve lies above the tangent, and 
at a point where the curve is concave downwards (as at C) the 
cxlrve lies below the tangent. At a point of inflection (as at B) 
the tangent evidently crosses the curve. 

• Points of inflection may also be defined as points where 

dhi d^ii 

(a) TT~ ^^'^ —^ changes sign, op 
oar* </a:« 

(b) — — • and — - changes sign. 
ay* dy* 

t It is assumed that fix) and/* (x) are continuous. The solution of Ex. 2, p. 137, shows how 
to discuss a case where f (x) and /"(x) are both infinite. Eyidently salient points (see p. 266; are 
excluded, since at such points /"(x) is discontinuous. 

136 



POINTS OF INFLECTION 137 

Following is a rule for finding points of inflection of the curve 
whose equation is y =/(x), including also directions for examining 
the curve in the neighborhood of such a point. 

First step. Findf'\x). 

Second step. Setf^\x) = Oj and solve the equation for real roots. 

Third step. Write f\x) in factor form. . 

Fourth step. Test /"(x) for values of a:, first a trifle less and then 
a trifle greater than each root found in the second step. If f\Q^ 
changes sign^ we have a point of inflection, 

Whenf'\x)^'\-j the curve is concave upwards v^+^.* 

Whenf\x) = — , the curve is concave downwards y"'^^^^. 

EXAMPLES 
Examine the following curves for points of inflection and direction of bending. 

1. y = 3a5*-4x« + l. 

Solution, f(x) = 3x*-4x« + l. 

First step. f'(x) = 36x2 - 24x. 

Second step. 36xa-24x = 0. 

.'. X = } and X = 0, critical values. 

Third step. f («) = 36 x (x ~ f). 

¥<mrth step. When x < 0, /" (x) = + ; and when x > 0, /" (x) = — . 

.*. curve is concave upwards to the left and concave downwards to the right of 
x = 0{Am figure). 

When x< J,/"(x) = -; and when x > i,/"(x) =+. 

.*. curve is concave downwards to the left and concave upwards to the right of 
X = J (B in figure). 

The curve is evidently concave upwards everywhere to the left of -4, concave 
downwards between A (0, 1) and B (}, i^), and concave upwards everywhere to 
the right of B. 

2. (y-2)« = (x-4). 

Solution, y = 2 + (X - 4)*. 

First step. ^ "^ S ^* " '*^" '' 

^-.?(^_4)-t. 
dx2 9^ ' 

• This may be easily remembered if we say that a vessel shaped like the curve where it is 
ooncaye upwards holds (+) water, and where it is concave downwards spills (-) water. 





138 DIFFERENTIAL CALCULUS 

Second step. When x = 4, both first and second deriyatives are infinite. 
Fourth step. Whenx<4, — ^ = + ; butwhenx>4, -i = -. 

dX* diXr 

We may therefore conclude that the tangent at (4, 2) is 

perpendicular to the axis of X, that to the left of (4, 2) the 

curve Lb concave upwards, and to the right of (4, 2) it is 

X concave downwards. Therefore (4, 2) must be considered 

a point of infiecUon. 

3. V = 2'* -^y^- Concave upwards everywhere. 

4. y = 5 — 2x — x^. Ans. Concave downwards everywhere. 

5. y = x". Ans. Concave downwards to the left and 

concave upwards to the right of (0, 0). 

6. y = x' — 8 x' — 9 X + 9. Ans. Concave downwards to the left and 

concave upwards to the right of (1, — 2). 

7. y = a + (x — by. Ans. Concave downwards to the left and 

concave upvrards to the right of (6, a). 

jS 

8. a*y = ax* + 2 a*. Ans. Concave downwards to the left and 

/ 4 ox 
concave upwards to the right of f a, — j • 

9. x« - 8 6x« + a«y = 0. Ans. Point of infiection is ( 6, — ^ J . 

10. y = X*. Ans. Concave upwards everywhere. 

11. y = x*-12x8 + 48x«-60. 

Ans. Concave upwards to the left of x = 2, 

concave downwards between x = 2 and x = 4, 
concave upwards to the right of x = 4. 

12. y = -• Ans. Concave downwards between ( ± —-, — \ 

x2 + 4a« VV32/ 

concave upwards outside of these points. 



13. y = x+36x2-2x»-x*. Ans. Points of infiection at x = 2 and - 3. 

x* / 9 a \ 

14. y = — . Ans. Concave upwards to the left of I — 8 a, ) 1 

x2 + 3a2 ^ \ 4 / 

concave downwards between T — 3 a, ) and (0, 0), 

concave upwards between (0, 0) and f 8 a, -j-)^ 
concave downwards to the right of f 8 a, — j • 

16. (-) + (~) = 1. Ans. Points of inflection are x = ± -^. 



POINTS OF INFLECTION 139 

16* oV = o^ - ^' -4iw. Points of inflection are x = ± - V27-3V38. 

6 

17. y = . Am. Points of inflection are x = 0, ±a VS. 

a« + x« 

18. y = sin z. Aia. Points of inflection are x = nr, n being any 

integer. 

19. y = tan x. Am. Points of inflection are x = rnr, n being any 

integer. 

20. y = x€-'. Am, x = 2 gives a point of inflection. 

21. Show that no conic section can have a point of inflection. 

22. Show that the graphs of e* and log x have no points of inflection. 

23. Show that the curve y (x^ + a') = x has three points of inflection lying on 
the straight line x — 4 €?y = 0. 

24. Show that the abscissas of the points of inflection of the curve y^ = /(x) 
satisfy the equation 

[/'(x)]» = 2/(x).r(x). 



CHAPTER XI 
DIFFERENTIALS 

99. Introduction. Thus far we have represented the deriyative 
of y =^f(x) by the notation 

We have taken special pains to impress on the student that the 
symbol dy 

dx 
was to be considered not as an ordinary fraction with rfy as numer- 
ator and dx as denominator, but as a single symbol denoting the 
limit of the quotient ^y 

Ax 
as Ax approaches the limit zero. 

Problems do occur, however, where it is very convenient to be 
able to give a meaning to dx and dy separately, and it is especially 
useful in applications of the Integral Calculus. How this may be 
done is explained in what follows. 

100. Definitions. lif{x) is the derivative otf{x) for a particular 
value of a;, and Ax is an arbitrarily chosen increment of x^ then the dif- 
ferential off(x), denoted by the symbol df{x)^ is defined by the equation 

{A) df{x)=f{x)Ax. 

If now f(x) = a;, then f^{x) = 1, and (A) reduces to 

dx = Ax J 

showing that when x is the independent variable the differential 
of X {= dx) is identical with Ax. Hence, if y =f{x), (A) may in 
general be written in the form 

(B) dy=zff(x)dx.* 

* On account of the position which the deriTatiTeyCa;) here oocnples, it \b Bometime« called 
the d\ff'erenticU co^cient. 

The student should obserre the important fact that, since dx may be given any arbitrary yalue 
whatever, dx is independent qfx. Hence dy is a function qf two independent vcuiables x and dx. 

140 



DIFFERENTIALS 



141 



The differential of a function equals its derivative multiplied by the 
differential of the independent variable. y 

Let us illustrate what this means geomet- 
rically. 

'Letf(z) be the derivative of y =/(a;) at P, 

Take dx = FQ, then 



dy=f{x)dx = tanr^FQ==^FQ=QT. 




M M'X 



Therefore rfy, or rff (a:), is the increment (= QT) of the ordinate 
of the tangent corresponding to dx.* 

This gives the following interpretation of the derivative as a 
fraction. 

If an arbitrarily chosen increment of the independent variable x 
for a point P (x^ y) on the curve y =f{x) be denoted by dx^ then in 
the derivative 



dx 



f (x) = tan T 



dy denotes the corresponding increment of the ordinate drawn to the 

tangent. 

101. da: and dy considered as infinitesimals. In the Differential 

Calculus we are usually concerned with the derivative, i.e. with 

the ratio of the differentials dy and dx. In some applications it 

is also useful to consider dx as an infinitesimal (see § 30, p. 21). 

Then by (iS), p. 140, and (2), p. 27, dy is also an infinitesimal. 

Hence in such cases dx and dy are corresponding variables each 

of which approaches the limit zero.f 

102. Derivative of the arc in rectangular 
coordinates* Let s be the lengthy of the arc 
AF measured from a fixed point A on the 
curve. Denoting the corresponding incre- 
ments by Ax^ Ay, As, we have from the figure 

(chord PQY=:(Axf^{Ay)\ 

* The student should note especially that thejdifferential {^dy) and the increment ("Ay) of 
the function corresponding to the same ralue dx{—Ax) are not in general equal. For, in the 
figure, dy =» QT but Ay = QP*. 

t In the Integral Calculus dx and dy are always regarded as infinitesimals. 

(Defined in §224. 




142 



DIFFERENTIAL CALCULUS 



Both multiplying and dividing the first member by (arc PQ)^ 
[=(A«)*], we get 

Dividing both members by (Ax)^ 

/ chord Pg \YA«Y_j^ /AyV 
\aKPQ ) \i!ix) ^ \iix) ' 

From this we get, when Ax approaches the limit zero, 

,1. , limit /chord P©\ ^ „ 
as8ummgthat^^^^(^— ^j = l. Hence 




(24) 



da 
doo 



--F^mf- 



Similarly, if we divide {A) by (Ay)" and pass to the limit, we get 
Also, from the above figure, 



cos 6 = 



Ax 



= ^._^^^,and 



chord P^ As chord P© 



chord P^ As chord Pg 

[Multiplying both numerator and denominator in each case by arc PQ (» A<).] 

As As approaches the limit zero, approaches the limit r, and the 
ratio of the arc P^ to the chord FQ approaches imity. Therefore 



(26) 



dx . dfi 

cos r = —- » sin r = -^* 
d8 da 



Using the notation of differentials, (24) and (25) may be written 



(98) 



-=[©■+']'•"• 



DIFFERENTIALS 



143 



An easy way to remember the relations (24) to (28) between the 
differentials dx^ dfy, ds is to note that they are correctly repre- 
sented by a right triangle whose hypotenuse 
is dsy sides dx and (2y, and angle at base r. 

Then , 

(i«=V(rfx)»4.(dy)«, 

and dividing by dx or rfy gives (24) or (25) 
respectively. Also, from the figure, 



/ 


X 


V 


dz 


^0 






X 



cosr = -T-> sin 
ds 



ds 



the same relations given by (26). 

103. Derivative of the arc in polar coordinates. In what follows 
we shall employ the same figure, notation, and reductions used on 
pp. 97, 98. From the right triangle PRQ 

(chord PQf = {PRf + (RQf 

= {p sin A^)* + (p + A/} — /} cos LOf 

= f? sin" A^ + (2 /) sin* ^ + Ap)". 

Multiplying and dividing the first 
member by (arc FQf [= (A»)^, and then 
dividing throughout by (A5)^ we get 




/ chord Pe y/A^Y^ , / sin Ag V 



)( 



\KtzPQ ) \LB) ^\ t^e 



) 



+ \ />8in 



A^ «^°-2 



A^ 
2 



^A^ 



Passing to the limit as A0 diminishes towards zero (see § 80, 
p. 98), we get 

(S)'= 



(») 



ds 
d$ 






In the notation of differentials this becomes 



m 



-^<i-*oi''- 



144 DIFFERENTIAL CALCULUS 

These relations between p and the differentials ds^ dp^ and d0 
are correctly represented by a right triangle whose hypotenuse is 
d9 and sides dp and pd0. Then 

ds = ^{pddf 4. (dp)\ 

and dividing by d6 gives (89). 

Denoting by -^ the angle between dp 
and ds, we get at once 

de 

which is the same as {A)^ p. 98. 

101 Formulas for finding the differentials of functions. Since . 
the differential of a function is its derivative multiplied by the 
differential of the independent variable, it follows at once that 
the formulas for finding differentials are the same as those for 
finding derivatives given in § 46, pp. 46-48, if we multiply each 
one by dx. 

This gives us 




I 


d!(c)=0. 




II 


d(x)=dx. 




III 


d(u + v — w)=du + dv — dw. 




IV 


d(cv)= cdv. 




V 


d(uv) = udv + »flJw. 




VI 


d (»,», ••■«,) = (v,v, •••»,) dv^ + («;,», . 


■■vj)dv,+ 




+ (»!"» • 


• '• »»-i) ^'v 


VII 


d(tf) = nif~^dv. 




Vila 


d(3f)= taf'^dx. 





VIII 



VIII a d 

c 






du — udv 



v" 
du 



Villi 



/c\ _ cdv 



DIFFERENTIALS 146 

IX <i(lOgat^) = 10g„«— • 

V 

X d (a*") = a* log a dv, 

Xa d {e'') = e" dv. 

XI d{u'') = vu^'^du + log w . w" . cZw, 

XII c2 (sin v) = cos t; dv. 

XIII d (cos t^) = — sin v dv. 

XIV * c2 (tan v) = sec* v rfv, etc. 

XIX c2(arc sin v) = =y etc. 

The term differentiation also includes the operation of finding 
differentials. 

In finding differentials, the easiest way is to find the derivative 
as usual, and then multiply the result by dop. 

Ex. 1. Find the differential of 

x + 3 
^ xa + 3 

_ (itg + 8)dx~(g + 3)2zda; _ {S-6x-z^)dz 

(x^ + 3)a "" (x» + 3)a ' "** 

Ex. 2. Find dy from 

iSoZutton. 2 b^xdx - 2 a^dy = 0. 

,\ dy = -—dx. An», 

Ex. 3. Find dp from 

ffl:=za^ cos 2 0, 

Solutum, 2 pdp = - a^sin 2 • 2 d0. 

a2 8in2^,^ 
.«. dp=: de. 

P 
Ex. 4. Find d [arc sin {St -4 <•)]. 

SchUum, d [arc am (3 1 — 4 <»)] = ' = - . ^n«. 

Vl-(3<-4e«)2 Vl - fi 



146 DIFFERENTIAL CALCULUS 

105. Successive differentials. As the differential of a function is 
in general also a function of the independent variable, we may 
deal with its differentiaL Consider the function 

d(df/) is called the second differential of y (or of the function) 

and is denoted by the symbol 

dY 

Similarly the third differential of y^ {i[(i(dfy)] is written 

and so on, to the nth differential of y^ 

dry. 

Since dx, the differential of the independent variable, is inde- 
pendent of X (see footnote, p. 140), it must be treated as a constant 
when differentiating with respect to x. Bearing this in mind 
we get very simple relations between successive differentials and 
successive derivatives. For 

dy=f{x)dx, 

and dry^r{x){dx)\ 

since dx is regarded as a constant. 

Also, d'y=f"{x)(dx)\ 

and in general d*y =/^">(a;) {dx)\ 

Dividing both sides of each expression by the power of dx 
occurring on the right, we get our ordinary derivative notation 

dj^^^ ^''^' dj^'^ <'^^'"' d^r"'^ ^''^' 
Ex. 1. Find the third differential of 

SoltUUm. dy = (6aj* - 6x« + 8)dx, 

cPy = (20a:«-12x)((to)2, 
d»y = (60 x» - 12) {dx)*, Ans. 

Note. This is evidently the derivative of the function multiplied by the cube of 
the differential of the independent variable. Dividing through by (dx)", we get the 
third derivative 

rL?^ = 00x2-12. 
dx* 



DIFFERENTIALS 147 



DifEerentiate the following, using diflerentials. 

1. y = ax«-ea« + cx + <l. dy = (8 ax« - 2 6x + c) das. 

2. y = 2a:*-8x'+6ari + 6. dy = (6«* - 2«'"* -6ar«)dx. 
8. y = (a2-x?)». dy = -10aj(a»-x«)*cte, 

4. y = Vl-fx«. dy = -r=L=dx. 

Vl-f «« 



(TT^* ^"(l + x«)-+ 



SaS^dx 

6. y = logVl-x». <'y = 2(xTi,i) ' 

7. y = (c* + c-*)». dy = 2(^*-e-««)d«- 

8. y = c*logx. dy = (?'(logx + -)dx. 






l + 8in0 

10. p = tan0 + 8ec^. ^""^cofi^S" 

11. r = itan»^ + tan^. dr = sec*M^. 

^, . , 8(logx)«dx 

12. fix) = (log x)». f(z)dx = -^-^ 

13. »(Q= . 0'(Od« = 



(1 - «»)« (1 - «^* 

dy 



16. d[arctanlogy]=: 



ydy 



16. d fr arc vers ?^ - V2 ry - y^l = —=0= 
L r J V 2 ry — y* 

17 dr-^5^-iiogtiui^i=--^- 




CHAPTER XII 
RATES 

106. The derivative considered as the ratio of two rates. Let 

y =/(^) 
be the equation of a curve generated 
by a moving point P. Its coordinates 
X and y may then be considered as 
functions of the time, as explained 
"Y in § 84, p. 104. Differentiating with 
respect to ^, by IIYI, we have 

At any instant the time rate of change of y (or the function) equals 
its derivative multiplied by the time rate of change of the independent 
variable. 

Or, write (81) in the form 

dp 

dt 

The derivative measures the ratio of the time rate of change of y 
to that of X, 

— being the time rate of change of length of arc, we have from 
(12), p. 105, 



WH^) 



which is the relation indicated by the above figure. 

Ex. 1. A point moves on the parabola 6 ^ = x^ in such a way that when x = 6 
the abecissa is increasing at the rate of 2 ft. per second. At what rates are the 
ordinate and length of arc increasing at the same instant ? 

148 



RATES 



149 



Solution. Differentiating with respect to the time t, we get 



dy 1 dx 

— = - . X — ' 

dt Z dt 



This means that at any point on the parabola 

(Rate of change of y) = (J z) {Rate of change of x). 

dx 
But from the problem, when x = 6, — = 2 ft. per sec. 

dt 

Substituting in (A), ^ = - . 0.2 ft. per sec. = 4 ft. per sec. 

dt 3 



P(M) 




Also, 



— = V4 + 16 = 2 Vs ft. per sec. 
dt 



That is, at the point P (6, 6) the ordinate changes in value twice as rapidly as 

the abscissa. 

dtf 
If we consider the point P' (— 6, 6) instead, the result is -^ = — 4 ft. per sec. 

dt 

The minus sign indicates that the ordinate is decreasing as the abscissa increases. 

Ex. 2. A man is walking at the rate of 5 miles per hour towards the foot of a 
tower 60 ft. high standing on a horizontal plane. At what rate is he approaching 

the top (a) at any instant ; (b) when he is 80 ft. from 
-f the foot of the tower ? 

I Solution, Let y = distance from top and x = dis- 

i. tance from foot of tower at any instant. Then 

f x* + (60)2 = y2. 

{ Differentiating with respect to the time t gives 




*._ 



dy 
dt 



xdx 
ydi 



(a) Hence he is approaching the top - times as fast as he is approaching the foot. 

dx 

(b) When x = 80, y = 100 ; and we have given — = 5 x 5280. Therefore 

dt 

• dt/ 80 

-^ = — X 5 X 5280 ft. per hour 
d« 100 *^ 

= 4 miles per hour. 

• 

Ex. 3. A circular plate of metal expands by heat so that its radius increases 
uniformly at the rate of .01 inch per second. At 
what rate is the surface increasing (a) at any instant ; 
(b) when the radius is 2 inches ? 

Solution. Let x = radius and y = area of the plate. 

Then y = irx^, and diflerentiating with respect to the 

time t, 

dy a dx 

dt dt 

(a) That is, at any instant the area of the plate is 
increasing in square inches 2 tx times as fast as the radius is increasing in linear 
inches. 




150 DIFFERENTIAL CALCULUS 

dx 
(b) When — = .01 in. per sec. and x = 2 in. 

dt 

dy 

-p = .04 r sq. in. per sec. ; 

i.e. the area is increasing .04 r sq. in. per sec. at that instant 

Ex. 4. An arc light is hung 12 ft. directly above a straight horizontal walk on 

which a man 5 ft. in height is walking. How fast Is the man^s shadow lengthening 

when he is walking away from the light at the rate of 108 ft per minute ? 

£, Solution. Let x = distance of man from a point directly 

under light X, and y = length of man*s shadow. From 

the figure, 

y : y + « : : 5 : 12, 

or, y = fx. 

Differentiating, -r^ = r -r- ; 

*' dt 7 dt 

Le. the shadow is lengthening f as fast as the man is walking, or 120 ft per minute. 




1. In a parabola y> = 12 x, if z Increases uniformly at the rate of 2 in. per second, 
at what rate is y increajsing when x = 3 in. ? Ans, 2 in. per sec. 

2. At what point on the parabola of the last example do the abscissa and ordi- 
nate increase at the same rate ? Ans. (8, 6). 

3. In the function y = 2 x' + 6, what is the value of x at the point where y 
increases 24 times as fast as x ? Ans. x = ^ 2. 

4. Find the value of x when the function 2 x' ~ 4 is decreasing 5 times as rapidly 
as X increases. Ara. x = ~ }. 

6. Find the values of x at the points where the rate of change of 

x»-12x2 + 46x-13 
is zero. Ans, x = 8 and 6. 

6. What is the value of x at the point where x* ~ GX^ + 17 x and x* — Sx change 
at the same rate ? Ana. x = 2. 

7. At what point on the ellipse 16x> + Oy' = 400 does y decrease at the same 
rate that x increases ? Ans, (3, Y)* 

8. Given y = x> — 6x3 + 8x + 5; find the points at which the rate of change of 
the ordinate is equal to the rate of change of the slope of the tangent to the curve. 

Ans, X a 1 and 5. 

9. Where in the first quadrant does the arc increase twice as fast as the sine ? 

Ans. At 60^. 



RATES ISl 

10. The side of an equJlatenl txiang^e is 84 inches long, and is increasing «t the 
rate of 3 inches per hoar ; how £ul is the area increasing ? 

Ans. 36 Vs sq. in« per hoar. 

11. Find the rate of change of the area of a aqnare when the side b is increas- 
ing at the rate of a units per second. Ans. 2 od sq. units per sec. 

12. (a) The volume of a spherical soap babble increases how many times as fast 
as the radius? (b) When its radius is 4 in. and increasing at the rate of f in. per 
second, how fast is the volume increasing ? « Ans, (a) 4 «t^ times as fast ; 

(b) 8S w cu. in. per sec 

13. One end of a ladder 50 ft. long is leaning against a perpendicular wall stand- 
ing on a horizontal plane. Supposing the foot of the ladder be pulled away from the 
wall at the rate of 3 ft per minute ; (a) how fast is the top of the ladder descending 
when the foot ia 14 ft. from the wall ? (b) when will the top and bottom of the 
ladder move at the same rate ? (c) when is the top of the ladder descending at the 
rate of 4 ft. per minute ? Ana, (a) | ft. per min. ; 

(b) when 26 V2 ft from wall; 

(c) when 40 ft. from wall. 

14. A barge whose deck is 12 ft. below the level of a dock is drawn up to it by 
means of a cable attached to a ring in the floor of the dock, the cable being hauled 
in by a windlass on deck at the rate of 8 ft per minute. How fast is the barge 
moving towards the dock when 16 ft away ? Ana, 10 ft. per minute. 

16. An elevated car is 40 ft. immediately above a surfi^e car, their tracks inter- 
secting at right angles. If the speed of the elevated car is 16 miles per hour and of 
the surface car 8 miles per hour, at what rate are the cars separating 6 minutes 
after they met ? An», 17.8 miles per hour. 

16. One ship was sailing sotith at the rate of 6 mUes per hour ; another east at 
the rate of 8 miles per hour. At 4 p.m. the second crossed the track of the first 
where the first was two hours before, (a) How was the distance between the ships 
changing at 3 p.m. ? (b) how at 6 p.m. ? (c) when was the distance between them 
not changing ? Ana. (a) Diminishing 2.8 miles per hour ; 

(b) increasing 8.73 miles per hour; 
. (c) 3.17 P.M. 

17. Assuming the volume of the wood in a tree to be proportional to the cube 
of its diameter, and that the latter increases uniformly year by year when growing, 
show that the rate of growth when the diameter ia 3 ft is 86 times as great as when 
the diameter ia 6 inches. 



CHAPTER XIII 
CHANGE OF VARIABLE 

107. Interchange of dependent and independent variables. It is 

sometimes desirable to transform an expression involving deriva- 
tives of y with respect to x into an equivalent expression involving 
instead derivatives of x with respect to y. Our examples will 
show that in many cases such a change transforms the given expres- 
sion into a much simpler one. Or, perhaps x is given as an explicit 

function of ^ in a problem and it is found more convenient to use 

- . . . , dx d^x ^ ^, . , . dy d^y ^ 

a formula mvolvmg -—, ^-r, etc., than one mvolving -^i -— ^, etc. 

ay djf ax cwr 

We shall now proceed to find the formulas necessary for making 
such transformations. 

Given y:=^f(^^ then from XXYII we have 

(88) ^ = JL, — :?fcO 

d,^ d'X cly 

dy 

giving -f- in terms of --• Also, by XIYI, 
ax dy 



(^) 



^ = i-/^\ = A/^\^, or, 
da? dx\dxj dy\dxj dx 

do? dy\dx\ dx 



\dy] \dy) dy 

Substituting these in {A)y we get 

(84) p.^^^ 

^dy) 
152 



CHANGE OF VARIABLE 163 

giving -z^ in terms of — and — • Similarly, 

^ ' dx^ /dx\* • 

\dy) 

and so on for higher derivatives. This transformation is called 
changing the independent variable from x to y, 

Ex. 1. Change the independent variable from x to y in the equation 
Schdion, Substituting from (33), (34), (36), 




dv \dv* 




a much simpler equation. 






dy* dy* 



106. Change of the dependent variable. Let 

{A) y=f{^)> 

and, suppose that at the same time y is a function of Zj say 

We may then express -^r-^ -7^» etc., in terms of -;-, — -r, etc., as 
- „ ax dxr dx dor 

follows. 

In general, ^e is a function of y by (-B), § 55, p. 57, and since y 

is a function of x by {A)^ it is evident that z is a function of x. 

Hence by IIVI we have 



154 DIFFERENTIAL CALCULUS 

Similarly for higher derivatiyes. This transformation is called 
changing the dependent variable from y to z^ the independent vari- 
able remaining x throughout. We will now illustrate this process ' 
by means of an example. 

Ex. 1. Having given the equation 

change the dependent variable from y to 2 by means of the relation 
(F) y = tan «. 

, Schition, From (F), 

-^ = 8ec*2-— I -4 = sec«z— - + 28ec»«tan«i — l . 
' dx dx dx^ dx^ \dx/ 

Substituting in (E), 

a ^2^ . o 4^* /<^\* -i . 2(l + tanz)/ - dz\* 
sec^z— - + 28ec*2;tana?i — ) =1+ — r^(8ec««~l, 

dx« \dx/ l + tan«« \ dxJ 

andreducing, weget— - — 2( -— ) =co8*«. Ana, 

dx^ \dx/ 

109. Change of the independent variable. Let y be a function of 
x^ and at the same time let x (and hence also g) be a function of a 
new variable t It is required to express 

dv d^y 
ax dor 
in terms of new derivatives having t as the independent variable. 

^ * dy __dg dx 

dt dx dt 
dy 

(A) ^ = ^. 

^ ' , doc dM_ 

at 

±/dy\ 

d^y _d /dy\ _ d /dy\ dt dt \dx) 

d7? dx\dx) dt\dx) dx dx 

di 
But differentiating (A) with respect to ^ 



dt\dx) dt 



ldy\ dxd^y dy d^x 



dt 
dx 

\Ttj 



dt df dt df 




CHANGE OF VARIABLE 



166 



Therefore 

{B) 



dap d*y dy dhp 
d^ _ dt dt* dt d^ , 

\dt) 



and so on for higher deriyatives. This transformation is called 
changing the independent variable from x to t It is usually better 
to work out examples by the methods illustrated above rather 
than by using the formulas deduced. 



Ex. 1. Change the independent variable from x to £ in the equation 



"^S-^^l--"' 



by means of the relation 

Solution. ' 

(S) 

Also, 



dx 

-— = e*. therefore 

dt ' 



dt 
dz 



— 0-t 



W 



dy dy dt ^, ^ 
-7^ = -^ — - , therefore 
dx dt dz 

^ = er'^. Also, 

dx dt 



dx* dxKdt/ dt dx dt\dt/dx dt dx 



Substituting in the last result from (E), 



dafi di^ dt ' 



Substituting (D), (F), (€F) in (C), 



= 0; 



and reducing, we get 



(2*1/ 

^ + y = 0. Ans. 



Since the formulas deduced in the Differential Calculus gener- 
ally involve derivatives of g with respect to x^ such formulas as 
(A) and (B) are especially useful when the parametric equations of 
a curve are given. Such examples were given on pp. 92-97, and 
many others will be employed in what follows. 



156 



DIFFERENTIAL CALCULUS 



110. Simultaneous change of both independent and dependent 
variables. It is often desirable to change both variables simul- 
taneously. An important case is that arising in the transfoimar 
tion from rectangular to polar coordinates. Since 

x=:p cos and y=^p sin 0, 
the equation 

becomes by substitution an equation between p and 6 defining p 
as a function of 0. Hence /?, 2:, y are all functions of 0. 

Ex. 1. Transform the formula for the radius of curvature (40), p. 103, 

[■-(l)T 



W 



2J = 






into polar coordinates. 

Solution, Since in (A) and (£), pp. 164, 155, t is any variable on which x and y 
depend, we may in this case let t = $, giving 

dy 



(5) 



{C) 



ox ox 

dz d*y _ dy cPx 
d^y _ ded^~^d^ 

\de) 



Substituting {B) and (C) in (A), we get 

\de) \dB/ ded0^ ddde» 



2J = 



( 



dx\a 
d$) 



/dx\« 

Kde) 



■» or. 



(D) 



U = 



dx d^y _^ dy d?x 
dSd^"" dSd^ 



But since x = p cos 9 and y = psin 6, we have 

— = — psm^ + cos^-f-; -^ = pcoB^ + 8in^~; 
d^ dtf d^ '^ dtf' 

^ = -pcos^-2sin^3^ + cos^^; ^ = - psin^ + 2co8^^ + sin^^. 
d^ dS d0^ de^ d$ dff^ 



CHANGE OF VARIABLE 167 

Substitating these in (D) and reducing, 

B= — — -. ^iw. 

'^ KdeJ *^dfi 

EXAMPLES 



Change the independent variable from x to y in the four following equations. 

dx\«n* 
dy. 



Mtn' . . [■+(!)*] 



1. J8 = Ara. i? = - 

dx« dya 

dx« Vdx/ dy« dy 

8. X— ^ + ( ~ ) - T^ = 0. Am. x^-:-H-(t-) =<>• 

dx« \dx/ dx dy'^ \dy/ 

\ dx )\djfi) \ dx Jdxdsfi Kdy^J \dy Jdy* 

Change the dependent yariable from y to z in the following equation. 

Change the independent yariable in the following eight equations. 

dx« 1 - x2 dx 1 - x« • dt^ '' 

7. (i_x«)^-x^ = 0, x = co8z. Am. ^ = 0. 
^ ^dx^ dz dz^ 

8. (l-y»)??-y?^ + o*M = 0, y = 8inx. ^n«. ^ + a^u = 0. 

dy* dy dx* 

dx* dx X* « dz* 

d*p -. - d*u du ^ d'v 

10. x»^ + 3x2^ + x5^ + r = 0, x = c«. Am. ^ + v = 0. 

dafi dx^ dx d^ 

11. ^ + .lfL*^ + _y_ = o, x = tan^. ^n«. ^ + y = 0. 
dx* 1 + X* dx (1 + x2)2 * de^ ^ 

12. h «u h sec* « = 0, « = arc tan t 

"^^ *• d*u du 

Am. (l + <*)^ + (2« + MarctanO — + 1 = 0. 
di* df 

13. x«?| + a«v = 0. » = 1. 4«.. ^-?^ + «ay = o. 

dx* z dz* z dz 



168 DIFFERENTIAL CALCULUS 

In the following four examples the eqaationa are given in parametric form. 

dv d^y 

Find -^ and —4 in each case. 
dx dx^ 

14. X = 7 + 1«, y = 3 + «a - 8t*. Arut, ^ = l-6««, ^ = -6. 

dx dx^ 

15. « = cote, y = 8in»«. Am. ^ = -3Bin*tco8e, !^ = 38in««(4 - 68in«t). 

dx dx* 



16. z = a(co8 1 + < sin t), y = a(8in t — t cos t). 



Ana, -^ = tan t, 



dx dafi atcoe^t 

17. x = ; — -» y = - — -• Ans, —- = — 1, --— = 0. 

l + tl + t dxdx* 

dy 

18. Transform — - by assuming x = p cos 6^ y = pmn0. 



Ans, 



19. Let /(x, y) = be the equation of a curve. Find an expression for its 

slope ( -^ ) in terms of polar codrdinates. ^ . , ^dp 

\dx/ ^ pcos^ + sm^^ 

A dy d$ 



— psm^ + costf-?- 
'^ d$ 



CHAPTER XIV 



CURVATURE. RADIUS OF CURVATURE 



111. Curvature. The shape of a curve depends very largely 
upon the rate at which the direction of the tangent changes as 
the point of contact describes the curve. This rate of change 
of direction is called curvature and is denoted by K. We now 
proceed to find its analytical expression, first for the simple case 
of the circle, and then for curves in general. 

112. Curvature of a circle. Consider a circle of radius E. Let 

r = angle that the tangent at P makes with OX, and 

r + At = angle made by the tangent at a neighboring point F'. 

Then we say 

At = total curvature of arc FF', 

If the point F with its tangent be 
supposed to move along the curve to 
P', the total curvature (= At) would 
measure the total change in direction, 
or rotation, of the tangent; or, what 
is the same thing, the total change in 
direction of the arc itself. Denoting by s the length of the arc 
of the curve measured from some fixed point (as A) to P, and 
by As the length of the arc FF^^ then the ratio 

At 

A« 
measures the average change in direction per unit length of arc.* 
Since from the figure. 




or, 



As=^B' At, 
A« b' 



* Thus, If At s. - radlanf (« a(n» ftnd a«b 3 oentimeten ; then 

At » ^ 

— s — radians per centimeter*- 10" per centimeter >- aTerage rate of change of direction. 

A« 18 

169 



160 



DIFFERENTIAL CALCULUS 



it is evident that this ratio is constant everywhere on the circle. 
This ratio is by definition the curvature of the circle^ and we have 



(86) 






The curvature of a circle equah the reciprocal of its radiu9. 

113. Curvature at a point. Consider any curve. As in the last 

section, 

At = total curvature of the arc FP\ 

At 

= average curvature of the arc FF\ 



and 



A^ 



More important, however, than the notion of the average curva- 
ture of an arc is that of curvature at 
a point This is obtained as follows. 
Imagine P' to approach P along the 
curve, then the limiting value of 

the average curvature ( = — j as P' 

• approaches P along the curve is defined 
as the curvature at P, i.e. 

Curvature at a point = . ^ ( -— ) =-r-' 

^ A« = \A8j d% 




(87) 



X = -—- = curvature. 
ds 



If we suppose P to move along the curve, t and s are functions 

of the time U and we may write 

dr 

j^_dT __dt 

dt 

dr 
where 3- = angular velocity of rotation of the tangent, 
dt 

and — = V = magnitude of the velocity of P in the path. 



Let t; = 1, then 



J5r= 



dr 
dt 



and we have the theorem : 

The curvature at P is equal to the angular velocity of the tangent 
at P when P describes the curve with unit velocity. 



CURVATURE 161 

114. Formulas for curvature. It is evident that if in the last 
section, instead of measuring the angles which the tangents made 
with OX, we had denoted by t and t + At the angles made by the 
tangents with any arbitrarily fixed line, the different steps would 
in no wise have been changed, and consequently the results are 
entirely independent of the system of coordinates used. How- 
ever, since the equations of the curves we shall consider are all 
given in either rectangular or polar coordinates, it is necessary to 
deduce formulas for K in terms of both. We have 



dy 
or, T = arc tan-p- 

ax 

Differentiating with respect to «, 

^(dy\ 
dr ds \dxj 



tanT = ^, §45, p. 44 






Therefore, substituting in (A), 

dr d^ 



and, since jr = -;-i we have 

d% 

(88) iST = '• 

['+(i)T 



by in 



" Hi) 

d^y dx 

dr da? ds ^ 

(A) ^= ^Tva ' ByllYI 

1 + 



B.t g4= :■..,. ■ By(«),p,142 

S 11 + 



162 DIFFERENTIAL CALCULUS 

If the equation of the curve be given in polar coordinates, 

K may be found by transforming* the above formula by means 

of the relations 

x^p cos d, y=zpsm 01 
giving 

(S9) X=- ^ ^^^ 



\f^{%n 



de 

Ex. 1. Find the curvature of the parabola ffi = ^pz at the upper end of the 
latus rectum. 

Solution. ^^2p.d«y^_2£^^_4^ 

dz y dx^ y* dx y* 

Substituting in (38), ^ , 



(y2 + 4 p2)* ' 
giving the curvature at any point. At the upper end of the latus rectum (p, 2p) 

(4p« + 4pa)* 16 V^p* 4 V2p 

Ex. 2. Find the curvature of the logarithmic spiral p^e^ &i any point 

SohUian. ^^a^ = ap; -^ = alki^ = aV 

d$ ^' d^ 

Substituting in (39), 

115. Radius of curvature. By analogy with the circle [see (86) 
p. 160], the radius of curvature of a curve at a point is defined as 
the reciprocal of the curvature of the curve at that point. Denot- 
ing the radius of curvature by By we have 

<> The details of this tranBformatlon were given in Ex. 1 on pp. 156, 167. 
t While in our work it is generally only the numerical value of K that is of importanoe, yet 
we can give a geometric metuiing to its sign. Throughout our work we have taken the positive 

sign of the radical v 1 + (-p) • Therefore A" will be positive or negative at the same time as 

-j^ 1 that is (§ 98t p. 136), according as the curve is concave upwards or concave downwards. 

t Hence the radius of curvature will have the same sign as the curvature, i.e. 4- or -> according 
as the curve is concave upwards or concave downwards. 



CURVATURE 168 



or, substituting the values of K from (38) and (89), 

[> + (i)T 



(40) JB = 






.= ['■^('^)T 



'•-'^+KI)" 



a 



« X 



Ex. 1. Find the radius of curvature at any point of the catenary y = - (e» + e <>) . 
Soitition. ^ = 2 («"-«"*) ; ^ = o~ (^ + ^"")- Substituting in (40), 

XX XX 4. a * 

e°H- e « e» 4- e « 

2a 2a 

If the equation of the curve is given in parametric form, find 
the first and second derivatives of y with respect to x from {A) 
and {B)y p. 154, 155, namely: 

dy 

dt 

dx d^y dy d^x 
dJ^y 'dide~'did^ ^ 

\dtj 
and then substitute the results in (40). 

Ex. 2. Find the radius of curvature x)f the cycloid 

X = a (« — sin t), 
y = a(l — cosQ. 

Solution. -— = a(l — cost), — = asint; 

dt dt 

— - = asmt, -4 = a cost 



<^ <ia. 



164 DIFFERENTIAL CALCULUS 

Substituting in (B) and (0), and then in (40), p. 163, we get 

dy smt cPy a (1 — cost) a cos t — a sin t a sin t _ 1 

5x~l -cost' dx^" a«(l-cost)» " a(l-co8t)« 

^^L Vl-cont/J ^_2aV2_2coBf. 4m. 



a 



a(l-co8 0« 

Note. From (6), p. 90, we get 

length of nonnal = y Vl + (^J = «(1 - «08 «) Vl + (i3^^ 

=a V2 — 2 cos f . 

Hence from a comparison of the last two results : 
At any point on the cycloid the length of the radius of curvature 
is twice the length of the normal. 

EXAMPLES 

1. Find the radius of curvature of the equilateral hyperbola xy = 12 at the 
point (3, 4). An8, R = ^. 

2. What is the curvature of y = x* — 4 a^ — 18 x' at the origin ? 

An8. K= -Z6, 

3. Find the radius of curvature of the ellipse iW -f a^ = aV^ (a) at any 
point, (b) at end of major axis, (c) at end of minor axis (a > h), 

o*o* a h 

4. What is the radius of curvature of the curve 16^^ = 4x* — x* (a) at (0, 1), 
(b) at (2, 0) ? An^ (a) iJ = 0, (b) iJ = 2. 

6. Find the curvature of the cubical parabola a^y = x*. _ . 

- _, oa*x 

Ana. K = 



(a*-f 9x*)* 

6. Get the radius of curvature in the semicubical parabola ay^ = z\ 

A^, ^^x*(4a + 9x)« 

6a 

7. Find the radius of curvature of the curve y = x* + 6 x* -f- 6 x at the origin. 

Ana, R = 22.506. 

8. Find the point on the parabola y^ = 8 x at which the radius of curvature 
isT^J. Ana. (}, 8). 

9. Find the curvature of the curvef - ) + ( - ) = 1 at the point (0, b). 

\a/ \b/ 86 

Ana. K = — 



CURVATURE 



166 



10. Determine the radius of curvature of the curve a^ = bx^ -f- ca^ at the 
origin. Aii», R = co. 



11. Shove that the radius of curvature of the witch y* = — at the vertex 



1 ^ 

••i- 



12. Find the radius of curvature of the curve y = log sec x at any point. 

Atis, 12 = sec X 

13. Find K at any point on the parabola x* -f y* = a*. Ans, K = 



2(x + y)' 
14. Find R at any point on the hypocycloid x^ + y' = a'. Aii^. 12 = 3 {axy)^ 

16. Find R at any point on the cycloid x = r arc vers - — V2 ry — y*. 

.4ns. 12 = 2 V2ry 

Find the radius of curvature of the following curves at any point. 



16. The circle p = a sin B, 

Yt. The spiral of Archimedes p = oB. 

18. The cardioid p = a (1 - cos ^ 

19. Thelemniscatep> = a3cos29. 

$ 

20. The parabola p = a sec' - . 

$ 

21. The curve p = a sin* - . 

22. The trisectrizp = 2acos0-a. 

23. The equilateral hyperbola p^cos 2 tf = a*. 



An8. R = - 
2 

2\l 



Ans, 12 = (e±^ 



p2 + 2a« 
Ana. 12 = fV2ap 

^ns. 12 = — 
3p 

$ 
Ans, JS = 2a8ec'- 

2 

Ans, 12 = }asin3~ 
^ns. j^^^(5-4cos^)t 



9 - 6 cos 9 

.4ns. 12 = ^ 
a2 



ox on. 1 a(l-c2) . -, a(l-e«)(l-2ccos^-f c*)* 
24. The conic p = —^ -• Ans, 12 = -^ f-^ ^ ^ 

1 — e cos ^ (1 — c cos ^)' 



26. The curve l*~of' « 

26. The hypocycloid j^~* 



X = acos'f, 
Sinn. 



Ans, 12 = 4(l+<2)2 
Ans, 12 = 3 a sin t cos t, 



27. The curve 
2a The curve 



Ans, R = at 



X = a (cost + tsinQ, 
y = a (sin t — t cos Q. 

Sx = a (m cos i + cos mt), 

m-l \ 2 / 



CHAPTER XV 




^b,o) 



THEOREM OF MEAN VALUE. INDETERMINATE FORMS 

116. Rolle's Theorem. Let y=f{x) be a continuous single-valued 
function of x vanishing f or x = a and a; = 6, and suppose that /'(x) 

changes continuously 
p when X varies from a 

to 6. The function will 
then be represented 
graphically by a contin- 
uous curve with a con- 
tinuously turning tan- 
gent as in the figure. 
Geometric intuition shows us at once that for at lecutt one value of x 
between a and b the tangent is parallel to the axis of X (as at P), 
i.e. the slope [=f'{x)] is zero. This illustrates Rolle's Theorem : 

Iff{x) vanishes when x = a and a; = 6, and y 
f{x) andf{x) are continuous for all values of x 
from x=za to x==b^ thenf{x) will be zero for 
at least one value of x between a and b,* 

This theorem is obviously true, because 
as X increases from a to 6, f{x) cannot always 
increase or always decrease as x increases, 

since f{a) = and f(b) = 0. Hence for 
at least one value of x between a and 6, 
f{x) must cease to increase and begin to 
decrease, or else cease to decrease and 
begin to increase ; and for that particular 
value of X the first derivative must be 
zero (§ 93, p. 118). 

* The second figure shows the graph of a function which is discontinuous (» w ) for x ■» c, a value 
lying between a and 6. The third figure shows the graph of a continuous function whose first 
derivatiye does not exist for such an intermediate value x = c. In each case it Is seen that at no 
point on the graph between x = a and x=b does the tangent (or curve) become parallel to OX, 

106 





THEOREM OF MEAN VALUE 



167 



117. The Theorem of Mean Value.* Consider the quantity Q 
defined by the equation 






or- 



ating ((7) we get 
Therefore, since 



Let F(x) be a function formed by replacing & by a; in. the left- 
hand member of (B); that is, 

( C) F{x) =f(x) -f{a) - (a: - a) Q. 

From (5), F(b)=0, and from ((7), F{a)= 0, 

therefore by RoUe's Theorem, p. 166, F\x) f must be zero for at 
least one value of x between a and 6, say x^. But by differenti- 

F^{x)==f{x)^Q. 

F\x^ = 0, then also/'(a;i) - © = 0, and 

Substituting this value of Q in (^), we get the Theorem of Mean 
Valuei 

(42) fm-f{a) ^ ^,^^^^^ a<x,<b 

o — a 

where in general all we know about Xy^ is that it lies between a and b. 
The Theorem of Mean Value can be easily interpreted geomet- 
rically. Let the curve in the figure be the locus of 

Take OC^a and OD=b; then 
f(a) = CA and f{b) = DB, giving 
AF=b-^a and FB =f{b) -f{a). 

Therefore the slope of the chord 
AB is ^ ^ f(b)^f(a) ^ 
AE b-a 

and the Theorem of Mean Value 
simply asserts that there is at least one point on the curve between 
A and B (as F) where the tangent (or curve) is parallel to the 
chord AB. If the abscissa of P is ar^, the slope at P is /'(Xj), and 

^« ^^« f(b)-f(a) _ .,. . 

l-a ~^ ^'>' 



-/W-/(«) 




Xx 



•Also called the Law qf the Mean. 



t If Fix) and F\x) are oontinuons. 



168 DIFFERENTIAL CALCULUS 

The student should draw curves to show that there may be more 
than one such point in the interval, and curves to illustrate on the 
other hand that the theorem may not be true iif(x) becomes discon- 
tinuous for any value of x between a and 6, or iif'(x) ceases to exist. 

Clearing (42) of fractions, we may also write the theorem in the form 

(48) f{h) = f(a) + (6 - «)/'(»!). 

Let 6 = a -I- Aa ; then b — a = Aa, and since a;^ is a number lying 
between a and 6, we may write 

2;^ = a -f ^ • Aa, 

where ^ is a positive proper fraction. Substituting in (48), we get 
another form of the Theorem of Mean Valuei 

(44) f(a + Aa) - /(a) = Aa/'(a + ff • Aa). < ^ < 1 

118. The Extended Theorem of Ifean Value.* Following the 
method of the last section, let E be defined by the equation 

(A) fib) -f{a) - (6 - a)f{a) - i (6 - aYE = 0. 

Let F(x) be a function formed by replacing 6 by a; in the left- 
hand member of (A); that is, 

(B) F{x)=f{x)-f{a) - (2; - a)f{a) - J (a: - a^E. 

From (A), i^(6)=0; and from (fi), F{a)=0, 
therefore, by RoUe's Theorem, p. 166, at least one value of x be- 
tween a and 6, say a:^, will cause F'(x) to vanish. Hence, since 

F'{x)=f'{x)--f(a)-{x-a)E, we get 

F\x,)=f\x,)^f{a) - {x, > a)E = 0. 

Since F^ (ar^) = and F^ (a) = 0, it is evident that F' (x) also satis- 
fies the conditions of RoUe's Theorem, so that its derivative^ namely 
F*'{x)j must vanish for at least one value of x between a and x^, 
say a;,, and therefore x^ also lies between a and 6. But 

F"{x)=f'{x)-E; therefore i^"(a;,)=/''(a;,)- ^ = 0, and 

E=r{x,). 

Substituting this result in (A)^ we get 

( 0) fib) =f(a) + {h- a)f (a) + ^{h- aff" (x,). a<x,<h 

•Also called the Extended Law cf the Mean. 



THEOREM OF MEAN VALUE 169 

In the same manner, if we define S by means of the equation 
f(P) -f(a) - (6 - a)f{a) - i (4 - o)«/"(a) - i (6 _ a)«5 = 0, 
we can derive the equation 

{!>) m ==f{a) + (6 - a)f (a) + i (i - a) V" (a) + i (6 - a)-/- (x,), 

where x^ lies between a and &. 

By continuing this process we get the general result,* 

+ ^^l~°rV -'>(«) + ^^^>'K). a < ^x < A 

|n — 1 In 

where x^.lies between a and b. (E) is called the Extended Theorem 
of Mean Value. 

119. Maxima and minima treated analytically. By making use 
of the results of the last two sections we can now give a general dis- 
cussion of maxima and minima of functions of a single independent 
variable. 

Given the function f{x). Let A be a positive number as small 
as we please ; then the definitions given in § 94, p. 118, may be 
stated as follows: 

If, for all values of x different from a in the interval [a — A, a + A], 

(-4) f{x) —f{a) =z= a negative number^ 

then f{x) is said to be a maximum when x = a. 
If, on the other hand, 

(B) f(x) — f{a) = a positive number^ 

then /(a;) is said to be a minimum when x = a. 
Consider the following cases: 

I. Letf(a)^0. 

From (48), p. 168, replacing 6 by a; and transposing /(a), 

( 0) fix) ^f(a) = {x-^ a)f (x,). a<x,<x 

• It Is aasamed that/(jr), f{x), fix), • • •,/(*) (x) exist throughont the interval [a, b]. 



170 DIFFERENTIAL CALCULUS 

Since /'(«) =^ 0, and f{x) is assumed as continuous, h may be 
chosen so small that f{x) will have the same sign as /'(a) for all 
values of x in the interval [a — h,a-\- A]. Therefore /'(a^J has the 
same sign as /'(a) (§§ 29-33). But x-^a changes sign according 
as a; is less or greater than a. Therefore, from ((7), the difference 

will also change sign, and by (A) and (£), /(a) will be neither a 
maximum nor a minimum. This result agrees with the discussion 
in § 94, where it was shown that /or all values of x for which f(x) is 
a maximum or a minimum the first derivative f (x) mu^t vanish. 

II. Letf{a)=0, and f (a) 4- 0. 

From ((7), p. 168, replacing J by x and transposing /(a), 

Since /"(a) i^ 0, and/"(z) is assumed as continuous, we may 
choose our interval [a — A, a + A] so small that/"(Xj) will have the 
same sign as /"(a) (§§ 29-33). Also {x — a)' does not change sign. 
Therefore the second member of (i>) will not change sign, and the 
difference ^, . ^, . 

will have the same sign for all values of x in the interval [a — A, 
a + A], and moreover this sign will be the same as the sign of f"{a). 
It therefore follows horn, our definitions {A) and {B) that 

{E) f(a) is a maximum iff '(a) = andf'\a) = a negative number; 
(F) f{a) is a minimum iff'{a) = andf"(a) = a positive number. 

These conditions are the same as (21) and (22), p. 124. 

III. Let f\a) =f"{a) = 0, and f\a) ^ 0. 

From (2>), p. 169, replacing 6 by 2; and transposing /(a), 

m f{x) -f{a) = 1^ (^ - «) V"(^a). a<x,<x 

As before, f"'{x^) will have the same sign as /'"(a). But (x — a)' 

changes its sign from — to + as 2: increases through a. Therefore 

the difference ^, . >,, . 

f{x) --f{a) 

must change sign, and f{a) is neither a maximum nor a minimum. 



THEOREM OF MEAN VALUE 171 

IV. Let f\a) ^f\a) = =/^-'>(a) = 0, and f^''\a) ^ 0. 

By continuing the process as illustrated in I, II, and III, it is 
seen that if the first derivative of /(a:) which does not vanish for 
2; = a is of even order (= n)^ then 

(45) /(a) is a maximum if/(>»)(a) = a negrative number; 

(46) /(a) is a minimum if /(«»)(a) = a positive number.* 

If the first derivative of /(a:) which does not vanish for a; = a is 
of odd order, then /(a) will be neither a maximum nor a minimum. 

Ex. 1. Examine x* — Qx^ + 24x — 7 for maximum and minimum values. 

Solution, f{z) = x« - 9x« + 24x - 7. 

/'(x) = 3xa-18x + 24. 

Solving 8x2 -18x + 24 = 

gives the critical values x = 2 and x = 4. .•. /(2) = 0, and / (4) = 0. 

Differentiating again, /"(x) = 6 x - 18. 

Since /''(2) = - 6, we know from (45) that/(2) = 13 is a maximum. 
Since /''(4) = + 6, we know from (46) that /(4) = 9 is a minimum. 

Ex. 2. Examine e' + 2 cosx + e-' for maximum and minimum values. 

Solution, f{x) = e*-f-2cosx + c-*, 

/'(x) = e« - 2 sin X - C-* = 0, for X = 0,t 
jT'lx) = c* - 2 cosx -f e-* = 0, for X = 0, 
/"'(x) = e»= + 2 sinx - C-* = 0, for X = 0, 
/i^(x) = e* -f 2 cosx -f e-* = 4, for X = 0. 

Hence, from (46), /(O) = 4 is a minimum. 

EXAMPLES 

Examine the following functions for maximum and minimum values, using 
method of the last section. 

1. 8x* — 4 X* -f 1. An8. X = 1 gives min. = ; 

X = gives neither. 

2. x» - 6x2 -f 12 X -f 48. Ana. x = 2 gives neither. 

3. (X - l)2(x + 1)». Am. X = 1 gives min. = ; 

X = I gives max. ; 
X = — 1 gives neither. 

4. Investigate 7fi — 5x* + 5x' — 1, at x = 1 and x = 3. 

5. Investigate x'* - 3x" + 3x -f 7, at x = 1. 

6. Show that if the first derivative of f(x) which does not vanish for x = a is 
of odd order (= n), then /(x) is an increasing or decreasing function when x = a, 
according as/<*)(a) is positive or negative. 

* As in §94, a critical ralae x=aiB found by placing the first derivatlye equal to zero and 
flolTing the resaltlng equation for real roots. 

t X "» is the only root of the equation e«- 2 sin x - e- •— 0. 



172 DIFFERENTIAL CALCULUS 

120. The Generalized Theorem of Mean Value. This theorem 
asserts about the two functions f{x) and F{x) that 

where x^ lies in the interval [a, 6] and F'(x) does not vanish in 
the interval. 

Proof. By multiplying both sides of (47) by F'{x^) and trans- 
posing /'(arj to the left-hand side, we see that this theorem requires 
that the equation 

F{b) - F{a) ^ ^ -^ ^ ^ 

shall have a root x^ lying between a and (. In order to make it 
possible to apply RoUe's Theorem, p. 166, let us try to construct a 
function having this left-hand member for a derivative and such 
that it (the function) vanishes for a; = a and x=b. Such a func- 
tion is 

/|5|^[i^(a:)-i^(a)]- [/(.)-/(«)], 

because it vanishes for x:=a and 2; =? 5, and hence by RoUe's 
Theorem its derivative must vanish for an intermediate value of 
2;, say x^i that is, 

F{b) - F{a) ^ ^""'^ ^ ^""'^ "' 

which is equivalent to (47). 

121. Indeterminate forms. When, for a particular value of the 
independent variable, a function takes on one of the forms 



— 1 ^1 0-QO, QO — 00, 0% 00**, 1*, 



it is said to be indeterminate^ and the function is not defined for 
that value of the independent variable by the given analytical 
expression. For example, suppose we have 

/(^) 
^(^) 

where for some value of the variable, as 2; = a, 

f(a) = 0, F(a) = 0. 



INDETERMINATE FORMS 



173 



For this value of x our function is not defined and we may 
therefore assign to it any value we please. It is evident from 
what has gone before (Case II, p. 28) that it is desirable to assign 
to the function a value that will make it continuous when 2; = a, 
whenever it is possible to do so. 

122. Evaluation of a function taking on an indeterminate form. If 
when 2; = a the function /(s;) assumes an indeterminate form, then 

M takm, as the value off{x) for x==a. 

The assumption of this limiting value makes f(x) continuous 
for x=z a. This agrees with the theorem under Case II, p. 23, 
and also with our practice in Chapter IV, where several functions 

assuming the indeterminate form ^ were evaluated. Thus, for 

2:= 2, the function — assumes the form -» but 

x — 2 



limit a:* — 4 



= 4. 



. a:=2 a:-2 

Hence 4 is taken as the value of the function for x = 2. Let us 
now illustrate graphically the fact that if we assume 4 as the value 
of the function for 2; = 2, then the function is continuous for 2; = 2. 

2:^-4 



Let 



y = 



2'-2 



This equation may also be written in the form 

y(2:-2) = (x-2)(2: + 2), 

or, (2;-2)(y-2:-2)=0. 

Placing each factor separately equal 
to zero, we have 

2; = 2, and y z= 2; -{- 2. 

In plotting, the loci of these equations 
are found to be the two lines AB and CD 
respectively. Since there are infinitely 
many points on the line AB having the 
abscissa 2, it is clear that when 2; = 2 
(= OM)^ the value of y (or the function) may be taken as any 




* The calculation of this limiting valae Is called evcUwUing the indeterminate form. 



174 DIFFERENTIAL CALCULUS 

number whatever, but when x is different from 2 it is seen from 
the graph of the function that the corresponding value of y (or the 
function) is always found from 

y = a; -I- 2, 

the equation of the line CD. Also, on CD, when a; = 2, we get 

y = jyrp = 4, 

which we saw was also the limiting value of y (or the function) 
f or a; = 2 ; and it is evident from geometrical considerations that 
if we assume 4 as the value of the function for a:= 2, then the 
function is continuous for 2; = 2. 

Similarly, several of the examples given in Chapter IV illustrate 
how the Umiting values of many functions assuming indeterminate 
forms may be found by employing suitable algebraic or trigonomet. 
ric transformations, and how in general these limiting values make 
the corresponding functions continuous at the points in question. 
The most general methods, however, for evaluating indeterminate 
forms depend on differentiation. 

123. Evaluation of the indeterminate form §• Given a function 

of the form ^^-^-^ such that /(a) = and Fia) = ; that is, the 

^{^) 

function takes on the indeterminate form - when a is substituted 

for z. It is then required to find 

limit f(x) 
x = a F(x) * 

Considering the functions f{x) and F{x) the same as in § 120 
and replacing 6 by a; in {iX)^ p. 172, we get 

F{x) F\x^) ' 

[Siii06/(a)= 0, and ^(a)= 0.] 

Since x^ lies between x and a, x^ approaches a as its limit when 
X approaches a, and we have 

limit f^ ^ limit f(x,) ^ limit f(x) ^, 
xz=a F{x) x^ = a F\x^ x=a F'{x) ' 

f*(x) 
* Assuming that ' does approach a limit as x approaches a. 



INDETERMINATE FORMS 175 

Iif'(z) divided by F\x) does not assume an indeterminate form 
for a: = a, we may write 

(i9i\ limit /(a?) _ /^(«) 

^^ » = «!?' (05) JP^a)' 

where f{a) = 0, i^(a) = 0, F\a) ^ 0. Hence 

Rule fqr evaluating the indeterminate form g* Differentiate the 

numerator for a new numerator and the denominator for a new 
denominator,* The value of this new fraction for the assigned 
valu£^ of the variable will he the limiting value of the original 
fraction. 

In case it so happens that 

f{a) = and F\a) = 0, 

that is, the first derivatives also vanish for*a;=a, then we still 
have the indeterminate form 


0' 

and the theorem can be applied anew to the ratio 

f{x) 

F\x) ' 
giving us 

limit /(^ ^ f\a) 
x=aF{x) F'\ay 

When also /"(a) = and F^\a)=: 0, we get in the same manner 

limit /(^ ^ f'\a) 
x=aF{x) F'^\a)' 
and so on. 

It may be necessary to repeat this process several times. 

* The student is warned against the very oareleaa but common mistake of differentiating the 
whole expression as a fraction by VIII. 

t If as «, the substitution x^- reduces the problem to the eyaluatlon of the limit for z« ; 
.^__ limit fix) Umit W^ limit w limit /W 

Therefore the rule holds in this case also. 



176 



DIFFERENTIAL CALCULUS 



^JLtU] _ 1-3 + 2 ' 
- X + iJ x=i~ 1 - 1 - 1 + 1 "" O' 





— 1 





T? 1 T? 1 * f(^) «*-3x + 2 

Ex.1. Evaluate ^ = -j^_^__^ when x = l. 

SoUUion. m= x»-3x.f2 
F(l) x»-x2 

/"(I) ^ '3aa-3 -| _ 3-3 _ 
F'(l) 3x2 - 2x - lJx=i~ 3 - 2 - 1 "" 
r(l) ^ 6x 1 ^ _6_ ^3 
1?'"(1) 6x-2Jx=i 6-2 2' 

Ex.2. Evalaate^^"^^^^"^''"^^. 

X = u X — sin X 

/(O) _e»-e-»-2x 
F(0) ~ X 

/^(O) __ C + g-» - 2 -1 _ 14 -1-2 _ 
2?^(0) 1-C08X Jx=o~ 1-1 ""O' 

F"(0) " sinx Jx=o~ ~0" '*•* 

r"(0) ^ e' + e- n 1 + 1 

-F'"(0) cosx 



indeterminate, 
indeterminate. 



Solution, 



n _ i-i-0 _ 
Jx=o~ 0-0 








Jx=0 1 



.*. indeterminate. 
. indeterminate, 
indeterminate. 
Ana, 



EXAMPLES 

Evaluate the following by differentiation.* 

J limit xg-16 
x = 4x2 + x-26 



2 limit x-1 
x = lx»-l' 

3 limit logx 
x = la._i' 

^ limit e*-g-* 
35 = *i sin X 

K limit tan X — X 
o. f. • 

X = u X — sin X 

Q limit log sin x 
' x = ^(T-2x)a' 

7 limit a'-b* 
^- x = 0--7~- 

g limit r» - gf^ - aV + a» 
' r = a »^ - a^ 



Ana. -' 


1 
n 



1. 

2. 

2. 
1 

^ ■ 

8 



1 A 
log-. 



Q limit ^ — arc sin $ 
* = sin8^ 



jQ limit sin X- sin 

X=0 x-0 

jj limit ey -f sin y ^ 1 
' y = log(l + y) 

io limit tan^-f sectf-1 
6' — tan ^ - sec ^ + 1 

23 limit 8ec20 -- 2tan0 
= ^ 1 + cos 4 



An^. — . 
6 



COS0. 



2. 



1 

— I 

2 



0. 



l^ limit gz-g^ ^ 

« = a a* - 2a»z + 2a2» - z*' ^'^ 

15. limit J^^e^)!^ 
x = 2(x^4)e' + e2x 



• After differentiating, the stadent shonld in every case reduce the resulting expression to its 
simplest possible form before substituting the value of the variable. 



INDETERMINATE FOiUVlS 177 

124. Evaluation of the indeterminate form §• 

In order to find ^"^'^ ^ 

x = a F{x) 

when i^^*^ fix) = 00 and i^™^^ Fix) = oo, 

that is, when for x^a the function 

fix) 
F{x) 
assumes the indeterminate form 

00 
00' 

we follow the same rule as that given on p. 175 for evaluating 
the indeterminate form -• Hence 

Rule for evaluating the indeterminate form ^ • Differentiate the 
numerator for a new numerator and the denominator for a new 
denmainator. The value of this new fraction for the assigned value 
of the variable will be the limiting value of the original fraction,* 

In case the new fraction is indeterminate for the given value of 
the variable, we repeat the process as in the last section. 

To prove this rule we must show that 

Umit /(£) ^ f'ja) 
x = a F{x) F\a) 

when ^"^l fix) = 00, I'^l Fix) = oo. 

First we assume that 

4:77-7 = U where Z is a definite number. 
x = a F\x) 

From the Generalized Theorem of Mean Value, p. 172, we have 
' F{x)-F(b) F'{x,) 1 ^ 

[Replacing h and ahj x and 6 respecttvely.] 

where b is an arbitrary number and F\x) does not vanish in the 
interval [a, a -f A]. From (A) 

or, fix) = m + ^ [F{x) - F(b)l 

* f'{x) and F\x) are aaBomed to be continuoua. 



178 



DIFFERENTIAL CALCULUS 



Dividing through by F(x), 



(^ 



F{x) F{x) F\x^) L F{x)\ 



In (£), let X approach a as a limit. Then 

limit /(^ ^ limit f{x^) 
x = a F{x) x=ia F\x^ ' 

Now let h approach the limit zero ; then x^ will approach the 
limit Oj and we get • 

limit f(x) ^ f(a) 
x=aF{x) F\a) 

In the same manner the rule may be shown to hold true when 

limit f{x) __ 



x=^a F\x) 



oo. 



Ex. 1. Evaluate -^- for x = 0. 



C8CX 



SohdioTL 



f(0) _\ogx 
F(0) 



caox 



-Ixi-O 



— CO 

■ ■ I 

CD 



.'. indeterminate. 



F'(0) 
F"(0) 



1 

X 



— cscxcotx, 
2Binxcosx 



Ll««o 



Bin^x 



XCOB 






.*. indetezminate. 



coax — xsinxJx-o 



] 







= — ^ = 0. ^fw. 



125. Evaluation of the indeterminate form 0*oo. If a function 
f{x) • <^ {x) takes on the indeterminate form * oo for rr = a, we write 
the given function 



<t>(x) 



m. 



00 
so as to cause it to take on one of the forms x or — , thus bringing 

it under § 123 or § 124. 



INDETERMINATE FORMS 179 



Ex. 1. Evaluate sec 8 x cos 6 x for x = - • 

2 

SobdUm, sec 8 X cos 5 x]x^ * = oo • 0. /. indeterminate. 

1 cos 6 X fisiS 

Substituting for sec 8x, the function becomes = ^^-^-i-. 

cob8x co8 8x F{x) 

f(-) 
\2/ cos6x-| , , , , ^ 

— = -— , = - • .-. indeterminate. 

/T\ cos3xJx«- 



F 



r(-) 

^ \2/ -sin5x.6-| 6 . 

^,(Z.\ -8m3x.8Jx,| 8 



126. Evaluation of the indeterminate form oo — oo. It is possible 
in general to transform the expression into a fraction which will 

assume either the form - or ^. 

00 

Ex. 1. Evaluate sec x — tan x for x = - . 

2 

SaMXoii, sec X — tan xlx»* = oo — oo. .*. indeterminate. 

By Trigonometry, secx-tanx = -i- - ?ilii^ = Ln^l^ = ZM. 

* ^ *" COSX COSX C08X F{^) 

Ai/ 1-sinxi 1-10 , , , , ^ 

— = . = ^ = - • .-. indeterminate. 

_/T\ COSX J*-? 

_,Vir\ — SinxJac-!^ —1 

EXAMPLES 

Evaluate the following expressions by differentiation.* 

y limit ax' + 6 . a limit tan^ -^^ , 

2 limit cotx 
x = Oiogx 

3 limit logx ^ fi' ^^'i j;;7^ ^• 

* X = 00 x* 

* In solving the remaining examples in this chapter It may be of assistance to the student to 
refer to J 37, pp. 36, 36, where many special forms not indeterminaie are evaluated. 



Ans. -• 
c 


4. 


limit tan^ 

^ = ^Un8^' 
2 


00. 
0. 


6. 


Umit V 2/ 
= - tan0 

2 



180 DIFFERENTIAL CALCULUS 

limit y 



6- - 



^n*0. 12. Mmur 2 _ 1 1 _1 



7. 



^^'"^i (X - 2 X) tan X. 2. 13. ^^\ T-l- _ _^1 . ^ i. 

x = - » = lLlogx logxJ 



8. "°^^^x8in2. a. 14. ^^^i [see ^ - tan ^]. 0. 



X=ao X tf = - 

2 



9. l^^^^o^-logx. [npcltive.] 0. 15. »™ij. [ 2 _ l_n 1 

x = ° = OL8in20 1-co8aJ 2 



,^ limit., * ^v oi. 1 i« limit ry_ 1 1 1 

10. ^ = r(l-tan^)8ec2(^. 1. ^^ y = 1 Lj^^ - i^J' a' 

' 11. >i°^iMa«-0«)tan^. i^'. 17. "°^^ f^ 1' - 

^ = a^" ''' 2a IT « = 0L4« 2«(c»« + l)J 8 

127. Evaluation of the indeterminate forms O^i 1", oo^ 
Given a function of the form 

In order that the function shall take on one of the above three 
forms, we must have for a certain value of x 

/(x)=0, <^(a:)=0, giving 0^ 

or, f{x) =1, «^ (a;) = 00, giving 1"; 

or, f{x) = 00, «^ {x) = 0, giving oo^ 

Let y=f{^y'''\ 

taking the logarithm of both sides, 

logy = «^(a;)log/(a:). 

In any of the above cases the logarithm of y (the function) will 
take on the indeterminate form 

Ooo. 

Evaluating this by the process illustrated in § 125 gives the 
limit of the logarithm of the function. This being equal to the 
logarithm of the limit of the function, the limit of the function 
is known.* 

• Thus, if Mmit logg y ■= a, then y - ««. 



INDETERMINATE FORMS 181 

Ex. 1. Eyaluate sc* when x = 0. 

Solution. This function aasames the indeterminate form (fi toTZ = 0. 



Let 


y = x*; 


then 


log y = X log X = • - 


By § 126, p. 179, 


, log X — a 




X 




1 



when X = 0. 



By § 124, p. 177, logy = — ^ = - x = 0, when x = 0. 

"x2 
Since y = x*, this gives loge x' = ; that is, x' = 1. Ans. 

I 
Ex. 2. Evaluate (1 + x)' when x = 0. 

Solution, This function assumes the indeterminate form 1* for x = 0. 

1 
Let y=:(l + x)^; 

then log y = - log{l + «) = oo • 0, when x = 0. 

X 

By § 126, p. 179, log y = ??iil±^ = ? , whenx = 0. 

X 



1 + X 1 

By § 123, p. 174, logy = -J-- = ; = 1, whenx = 0. 

1 1 + X 

a 11 

Since y = (1 + «)*, this gives log. (1 + x)* = 1 ; i.e. (1 + x)* = c. Ana. 



Ex. 8. Evaluate (cot x)»*»* for x = 0. 

SohUUm. This function assumes the indeterminate form ooP f or x = 0. 
Let y = (cot x)"*»* ; 

then log y = sin X log cot x = • oo, when x = 0. 

By § 126, p. 179, logy = ?2«5?^ = ^, when x = 0. 

cscx * 

— csc^x 

cot X sin x 

By § 124, p. 177, logy = > = — — = 0, when x = 0. 

— cscx cot X cos'x 

Since y = (cot x)«*n*, this gives log, (cot x)*^* = ; i.e. (cot x)«*»' = 1. Ana. 



182 



DIFFERENTIAL CALCULUS 



EXAMPLES 

Evaluate the following ezpressionB by diflerentiation. 



X = 1 



«• i'nc^r" 



3. "™^i(sin^>t"«. 

^ 2 



y = (X>\ ^ yj 



e 



1. 



1. 



&*. 



9. J^^ Q (cos m^)»*. 



1 

e 

c". 

1 

e 



CHAPTER XVI 

CIRCLE OF CURVATURE. CENTER OF CURVATURE 

128. Circle of curvature.* Center of curvature. If a circle be 
drawn through three points P^, P^, P, on a plane curve, and if P^ 
and P, be made to approach P^ along the curve as a limiting posi- 
tion, then the circle will in general approach in magnitude and 
position a limiting circle called the circle of curvature of the curve 
at the point P^. The center of this circle is called the center of 
curvature. 

Let the equation of the curve be 

(1) y=/(^); 

and let rr^, z^, x^ be the abscissas of the 
points Po, Pj, Pa respectively, (a', yS') 
the coordinates of the center, and -B' 
the radius of the circle passing through 
the three points. Then the equation 
of the circle is 

and since the coordinates of the points P^, P^, P, must satisfy this 
equation, we have 

({X, - a'y + (y, - I3y - 22" = 0, 
(2) .(x,^ar + {i/,-l3r^E^ = 0, 

Now consider the function of x defined by 

F{x)={x^ay + (tf - I3y - B'^, 

in which y has been replaced hyf{x) from (1). 
Then from equations (2) we get 

F{x,)=0, P(a:,) = 0, F{x,)=0. 




F,C«i.1'«> 



FiCpCi'l^i^ 



• Sometimes called the oiculating circle* 

183 



184 DIFFERENTIAL CALCULUS 

Hence by RoUe's Theorem, p. 166, F^(x) must vanish for at 
least two values of :r, one lying between x^ and a?^, say rr', and the 
other lying between x^ and a^j, say x" ; that is, 

F\x)=0, F'{x')=0, 

Again, for the same reason, F"(x) must vanish for some value 
of X between x' and a?'', say x^ ; hence 

F'\x,)=0. 

Therefore the elements a\ yS', B' of the circle passing through 
the points Pq, Pj, Pj must satisfy the three equations 

F(x,)=0, F'{x')=0, F"{x,)=0. 

Now let the points P^ and Pj approach P^ as a limiting position ; 

then Xy^y x^, x\ x'\ x^ will all approach a;,, as a limit, and the elements 

a, 13, R of the osculating circle are therefore determined by the 

three equations 

P(x,)=0, P'(a:o)=0, P"(Xo)=0; 

or, dropping the subscripts, what is the same thing, 

{A) (x-a)»-h(y-/3)» = i?, 

(P) (a: - a) + (y - /S) -/ = 0, differentiating {A). 

dx 

^^^ ^ "^ (^)' "^ ^^ ■" '^^ S " ^' differentiating (P). 

Solving (P) and ((7) for a: — a and y — yS, we get, {-^^^y 

dx [_ \dx) J 



"a;— a = 



(-») 



y-/8 



MS 

d^y 



d7? 

hence the coordinates of the center of 



curvature are 




P 



S [' + (S)T . . > + (S)" /^v 






(S-) 



CIRCLE AND CENTER OF CURVATURE 



185 



Sulistituting the values of x — a and y — /3 from (J)) in (J.), and 
solving for B, we get 



B = ± 



[l + /^^Yl' 



\dxj J 
d7? 



which is identical with (40), p. 163. Hence 

Theorem. The radiu9 of the circle of curvature equ(d$ the radius 
of curvature. 

From (23), p. 136, we know that at a point of inflection (as Q 
in the figure) 



da^ 



= 0. 



Therefore, by (88), p. 161, the curvature 
Jr= 0; and from (40), p. 163, and (Jg?), p. 184, 
we see that in general a, y3, B increase with- 
out limit as the second derivative approaches 
zero. That is, if we suppose P with its tangent 
to move along the curve to P', at the point of 
inflection Q the curvature is zero, the rotation 
of the tangent is momentarily arrested, and as the direction of 
rotation changes, the center of curvature moves out indefinitely 
and the radius of curvature becomes infinite. 





Ex. 1. Find the coordinates of the center of curvature of the parabola 2^ = 4 px 
corresponding (a) to any point on the curve; (b) to the vertex. 

Sol^ion. ^ = !£;^ = _1^. 
dz y dx^ y* 

(a) Substituting in (E), 

. y' + 4pa 2p y« ^o, . «« 

'4jpa" 



/3 = y- 



ya 4jpa 4p« 

Therefore ( 3 a; + 2 p, — =— ) is the center of curva- 
\ 4pg/ 

ture corresponding to any point on the curve. 

(b) (2 Pf 0) is the center of curvature corresponding 
to the vertex (0, 0). 



186 



DIFFERENTIAL CALCULUS 



129. Center of curvature the limiting position of the intersection 
of normals at neighboring points. 
Let the equation of a curve be 



(A) 



y =/(^)- 



The equations of the nonnaLs to the curve at two neighboring 
points Pj and Pj are* 

If the normals intersect at C\a\ 13% 
(?vVi) the coordinates of this point must 
satisfy both equations, giving 




-^CaCo'Vo) 



(B) 






Now consider the function of x defined by 

in which y has been replaced hyf{x) from (A), 
Then equations (B) show that 

<^(r,)=0, <l>{x,)^0. 

But then by Rolle's Theorem, p. 166, <^'(a:) must vanish for some 
value of X between Xq and x^^ say 2/. Therefore a' and yS' are deter- 
mined by the two equations 

<l>{x,)=0, <^'(a:')=0. 

If now Pj approaches Pq as a limiting position, then x* approaches 
x^, giving 

<l>{x,)==0, <l>\x,)=0; 

and C" (a', /3^ will approach as a limiting position the center of 



• From (2), p. 90, X and Y being the variable coordinates. 



CIRCLE AliD CENTER OF CURVATURE 187 

curvature C (a, 13) corresponding to F^ on the curve. For, if we 
drop the subscripts and write the last two equations in the form 

(»-«')+(y-^')f =». 

it is evident that solving for a* and yS' will give the same results 
as solving (B) and (C7), p. 184, for a and ;8. Hence 

Theorem. The center of curvature C corresponding to a point P 
on a curve is the limiting position of the intersection of the normal to 
the curve at P with a neighboring normal, 

130. Evolutes. The locus of the centers of curvature of a given 
curve is called the evolute of that curve. Consider the circle of 
curvature corresponding to a point P on 
a curve. If P moves along the given 
curve, we may suppose the correspond- 
ing circle of curvature to roll along the 
curve with it, its radius varying so as ^^«55>s^r:::^*"-^^^^ 
to be always equal to the radius of cur- />cV~'"A, / 

vature of the curve at the point P. The ( >» K ^ '1 ./o 

curve CCj described by the center of the V ^\ \\J/Pt 

circle is the evolute of PP^. ^-O,.*^ 

Formula (E)^ p. 184, gives the coordi- ■'' 

nates of any point (a, 13) on the evolute expressed in terms of the 
coordinates of the corresponding point (a-, y) of the given curve. 
But y is a function of 2;, therefore 

['<g)']^ , '-m 

a=^z-i= i^ — ^-=i — , /3 = yH ^ — ^ 

d^ ^ ^ 

dJ dx" 

give us at once the parametric equations of the evolute in terms of 
the parameter x. 

To find the ordinary rectangular equation of the evolute we 
eliminate x between the two expressions. No general process of 
elimination can be given that will apply in all cases, the method 



188 



DIFFERENTIAL CALCULUS 



to be adopted depending on the form of the given equation. In a 
large number of cases, however, the student can find the rectan- 
gular equation of the evolute by taking the following steps. 

General directions for finding the evolute. 

First step. Find a and fifrom (JB), p, 184. 

Second step. Solve the two resulting equations for x and y in 
terms of a and 13, 

Third step. Svhstitvte these values of x and y in the given equa- 
tion. This gives a relation between the variables a and fi which is 
the equation of the evolute. 

Ex. 1. Find the equation of the evolute of the parabola of y^ = 4px. 

dy 2p d^y 4p2 




SolvJLwn, 
First step. 
Second step. 
Third step. 






dx y dx^ y' 

o = 3x + 2jp, /3 = - 



4p3 



p/3a=l(a-2p)«. 



Remembering that a denotes the abscissa and /3 the 
ordinate of a rectangular system of co5rdinates, we see 
that the evolute of the parabola AOB is the semicubical parabola DC'E\ the 
centers of curvature for O, P, Pi, Pj being at C, C, Ci, C% respectively. 

Ex. 2. Find the equation of the evolute of the ellipse If^x^ + aV = a^^- 

Solution. -^ = > — - = 

dz a^y dx^ oV 



First step. 



Second step. 



*= a* ' 
(a^-l^)y^ 




Third step. (oa)* + (hfi)^ = (a* - 62)', 

the equation of the evolute EUE'B' of the 
ellipse ABA'R. E, W^ H\ H are the centers of curvature corresponding to the 
points Ay A\ Bj Bf on the curve, and C, C, C" correspond to the points P, P*, P", 



CIRCLE AND CENTER OF CURVATURE 



189 



When the equations of the curve are given in parametric form, 

we proceed to find -~ and •--^, as on pp. 154, 156, from 

ax dor 



(^) 



{B) 



dy 
rfy __ dt 
dx dx 

di 

dx d?y dy d?x 
^y Jdt'dj^'"dt'd^ ^ 

d3? "" y^Y ' 



and then substitute the results in formulas (27), p. 184. This 
gives the parametric equations of the evolute in terms of the same 
parameter that occurs in the given equations. 

Ex. 3. Find the parametric equations of the evolute of the cycloid. 

|x = o(<-BinQ, 
^ ' |y = a(l — cost). 

Sohdion, As in Ex. 2, p. 163, we get 

dy Bxnt d^ 1 



dx 1-cost dx* a(l-co8Q« 




Substituting these results in for- 
mulas (E), p. 184, we get 

im (« = «(* + sin <), 

^ ' (/3 = -a(l-cos<). Ana. 

Note. If we eliminate t between 
equations (2>) there results the rectan- 
gular equation of the evolute 0(y(f 
referred to the axes (Xa and (X/3. 
The co5rdinates of with respect to 
these axes are (— ira, —2a). Let 
us transform equations (D) to the new set of axes OX and OY, Then 

= 0: — ira, /3 = y-2a, « = f-T. 
Substituting in (D) and reducing, the equations of the evolute become 

fx = a(f — sinf), 
y = a (1 — cos t^. 

Since (J?) and (C7) are identical in form, we have : 

Tht eoolute of a cycloid is itself a cycloid whose generating circle equals that of 
the given cycloid. 



{E) 



190 DIFFERENTIAL CALCULUS 

131. Properties of the evolute. Differentiating a and 13 from (£7), 
p. 184, with respect to x gives 



(^) 



(^ 



da _ dy dx \d3?) d:f \^^) ^^ 
dx dx /^S[^* 

dl3 dx\d7?) da? \dxj da? 



Dividing (5) by (A), member for member, 

^ ' da di 

dx 

JO 

But -^ = tan t' = slope of tangent to the evolute at Cy and 
da 

-^ = tan T = slope of tangent to the given curve at the 

corresponding point P. 

Substituting the last two results 
in ((7), we get 

tan t' = . 

tanr 

Since the slope of one tangent is 
the negative reciprocal of the slope 
of the other, they are perpendicular. 
But a line perpendicular to the tan- 
gent at P is a normal to the curve. Hence 

A normal to the given curve is a tangent to its evolute. 
From (12), p. 105, and (A) and (jB), we have for an arc « of 
the evolute 




L \dx) J I /d^jA 






dx'J 



CIRCLE AKD CENTER OF CURVATURE 191 

But tbia is identically the result we get bj differentiating B, (|0), 
p. 163, with respect to z and then squaring. Therefore 

\dxj \dx) ' 

or, A» = ± an. 

That is, the radius of curvature of the given curve increoBes or 
deereageg at fait aa the arc of ike evoltUe increaeet. In our figure 
this means that p(,^ _ pf,^ ^^^ ^^^ 

The length of an arc of the evolute ia equal to the difference between 
the radii of curvature of the given curve which are tangent to thit arc 
at ita extremities.* 

Thus in Ex. S, p. 189, we observe that if we fold Q'P' (= 4 a) 
over to the left on the evolute, F" will reach to 0\ and we have : 

The length of one arc of the cydoid (at OO'Q^ ia eight timea the 
length of the radiua of the generating circle. 

133. Involutes and their mechanicAl construction. Let a flexible 
ruler be beat in the form of the curve C,C^ the evolute of the 
curve P,P,, and suppose a string 
of length B, with one end fastened 
at C, to be wrapped around the ruler 
(or curve). It is clear from the 
results of the last section that when 
the string is unwound and kept 
taut the free end will describe the 
curveJ*,i*,. Hencethenameeuo?«(e. 

The curve P,P, is said to be an 
involute of (7,C,. Obviously any 
point on the string will describe 
an involute, so that a given curve 
has an infinite number of involutes but only one evolute. 

The involutes P,P„ Pi'^t'i P"Pt' are called parallel curves since 
the distance between any two of them measured along their com- 
mon normals is constant. 

The student should observe how the parabola and ellipse on 
p. 188 may be constructed in this way from their evolutes. 

■ It I* iMDmed that -— <lo«a Dot otaugs ilgn. 



192 DIFFERENTIAL CALCULUS 

EXAMPLES 

Find the coordinates of the center of curvature and the equation of the evolute 
of each of the following curves. 

1. The hyperbola — — ~ = 1. Ans, a = ^ — - — '- — , /3 = — ^ — — — '-^ - 

a* cr a* 6* 

evolute (da)* - (6/3)' = (a^ + 6>)'. 

2. The hjrpocycloid x* + y* = a*. Ans. o = x + 3x*y*, /3 = y + 3 x'y* ; 

evolute (tt + ft* + (o - /3)' = 2 a'. 

8. The cycloid x = r arcvers =- — V2 ry - y^. 

iln«. tt = x + 2 V2 ry — y«, /3 = — y; 

evolute o = r arcvers ( - -) + V- 2r/3 - /S^. 
4. The semicubical parabola x^ = ay^. /v « ' r 

evolute 720 a/J^ = 16 [2 a + Va^ ~ 18 ao]« [ Va* - 18 oa - a]. 
6. The tractrix X = a log ^^^-^^^^ — - Va^ - y^. 

y 

a + Va^^r^ a« , a( - --\ 

Am. a = alog =^» /3 = —; evolute /3 = -\c« + e «/. 

a - -- V 5 _* 

6. The catenary y = -(ef^ ■\- e «). ^n«. o = x — -(c« — c «), /3 = 2y; 

2 2 

evolute o = a log ^ ^ (^ ""* ^ ) :p A (sa _ 4 ^s)*, 

2a 4a 

7. Fhid the coordinates of the center of curvature of the cubical parabola 
y* = a%;. a* + 16y* ^ a*y~9y« 

6a2y "^ 2 a* 

8. Show that in the parabola x^ + y^ = a* we have the relation a + /9 = 3 (x + y). 

9. Find the equation of the evolute of the cissoid y^ = 



2a ~x 
Am. 4006a«a + 1152a«/3a + 27/5* = 0. 

10. Given the equation of the equilateral hyperbola 2xy = a^; show that 

- , o_ (y + g)' ^ o _ (y - g)' 

From this derive the equation of the evolute (a + /9)' - (a — ft' = 2 a*. 
Find the parametric equations of the evolutes of the following curves in terms 
of the parameter t. 

12. The curve * = ^f' ^««. « = *<! + 2''-'*). 

18. The curve |* = "/^^ +'""«|' Ar^. j » = " f /' 

(y =a(8int-<cost). (/8 = a8in«. 



CHAPTER XVII 



PARTIAL DIFFERENTIATION 

133. Continuous functions of two or more independent variables. 
A function /(a:, y) of two independent variables x and y is defined 
as continuous for the values (a, b) of {Xy y) when 

limit 

25 = a/(^ y) =/(«i 6)» 

y = h 
no matter in what way x and y approach their respective limits a 

and J. This definition is sometimes roughly summed up in the 
statement that a very small change in one or both of the independ- 
ent variables shall produce a very small change in the value of the 
function.^ 

We may illustrate this geometrically by considering the surface 
represented by the equation 

Consider a fixed point P on the surface where a: = a and y = i. 

Denote by Aa; and Ay the increments of the independent vari- 
ables X and y, and by Lz the corresponding increment of the 
dependent variable 2, the coordinates of P' 

^"^ (a; 4. Aa:, y -f Ay, 2 4- A2). 

At P the value of the function is 

z =:f(a, b) = MP. 

If the function is continuous at P, then how- ^ 
ever Aa; and Ay may approach the limit zero, 
Az will also approach the limit zero. That is, M'P' will approach 
coincidence with MP^ the point P' approaching the point P on the 
surface from any direction whatever. 

* This will be better imdentood if the student again reads oyer § 33, p. 22, on oontinnoufl 
functions of a single yariable. 

103 




194 DIFFERENTIAL CALCULUS 

A similar definition holds for a continuous function of more 
than two independent variables. 

In what follows, only values of the independent variables are 
considered for which a function is continuous. 

134. Partial derivatives. Since x and y are independent in 

X may be supposed to vary while y remains constant, or the reverse. 
The derivative of z with respect to x when x varies and y remains 

constant* is called the 'partial derivative of z with respect to x^ and 

dz 
is denoted by the symbol — . We may then write 

dx 

... dz_ limit V f(x + i^y) -fix, y) -| 

^^> a;-Aa; = 0L Ai J' 

Similarly, when x remains constant* and tf varies, the partial 
derivative of z with, respect to y ia 

,Ji\ ?£_ limit r fjr, y + Ay) -f(x, y) ") 

^^> 8y-Ay = 0L Ay J* 

— is also written — f(x^ y) or — ; similarly 
dx dx*^ ^ ^^ dx '' 

— is also written — /(ar, y) or — . 
dy dy' ^ ' ^^ dy 

In order to avoid confusion the round ^f has been genei*ally 
adopted to indicate partial differentiation. Other notations, how- 
Bver, which are in use are 

(di)' \£) ' ^^^^' y)'-^/(^' y) ' f-^^' y)Jy{^^ y)y ^x/ ^J\ «x» V 

Our notation may be extended to a function of any number of 
independent variables. Thus, if 

u = F{x, y, z), 

then we have the three partial derivatives 

du du du^ dF dF dF 
— -, — , ——J or — — , — ) — • 

dx dy dz dx dy dz 

* The constant valaes are eubstitated in the function before differentiating. 
t Introduced by Jacobi (1801-1851). 



PARTIAL DIFFERENTIATION 195 

Ex. 1. Find the partial derivatives of « = ox^ + 2 bxy + c^. 

dz * 

Solution. — =z2ax + 2by, treating ]^ as a constant, 

dx 

— = 2 to + 2 c^, treating x as a constant. 
dy 

Ex. 2. Find the partial derivatives of u = sin (ox -{-by -{- cz). 

Solution, — = a cos (ax + by + cz)^ ti-eating y and z as constants, 

dx 

— = b cos (ox + 6y + cz), treating x and z as constants, 

a-, 

— = c COS lax + &v + <72)) treating y and x as constants. 
dz 

Again turning to the function 

we have by (A) defined — as the limit of the ratio of the increment 

dx 

of the function (y being constant) to the increment of a:, as the incre- 

dz 
ment of x approaches the limit zero. Similarly (B) has defined — • 

It is evident, however, that if we look upon these partial deriva- 
tives from the point of view of § 106, p. 148, then 

dz 

dx 

may be considered as the ratio of the time rates of change of z and 
X when y is constant, and o^ 

dy 

as the ratio of the time rates of change of z and y when x is 
constant. 

135. Partial derivatives interpreted geometrically. Let the equa- 
tion of the surface shown in the figure (next page) be 

z =f(x, y). 

Pass a plane EFGH through the point P (where a; = a and y = J) 
on the surface parallel to the XOZ plane. Since the equation of 
this plane is «/ = 6 

the equation of the section JPK cut out of the surface is 

z==f{x, 6), 



196 DIFFERENTIAL CALCULUS 

if we consider EF as the axis of Z and EH as the axis of X. lu 
this plane — means the same aa 

dz ^ 

-=-. and we hare 
dx 

— = tan MTP 
dx 

= slope of lection JK at P. 

Similarly, if we pass the plane 
BCD through P parallel to the 
YOZ plane, its equation is 



and for the section DPI, ~ means the same as --• Hence 
di/ dy 

~ = tan MT'P = slope of section 2)1 at P. 



Ex. 1. Qifen the ellipaoid h --- + — = 1 ; fln<i the slope of the section of 

the ellipaoid made (a) by the pliuie ]/ = 1 at, the point where z = 4 and z ii poritive; 
(b) by the plane z = 2 at the point where y = S and z is positive. 

Solution. Conddering y as constant, 

2i 2^ Si _ ^^ _ x_ 

24 ~0 ex~ ' °'' tx~~ iz' 

When z is constant, ?i^ + ^ ^ = 0, or, ^ = _ J^. 
12 Q di/ By 2z 

(a) Wheni/=landi = 4,i = -J^. ..^^-.-J^. Am. 

(b) Whena! = 2andi* = 3,j: = — . .•. — = ~-Vi. Am. 

-ft iy 1 



^■Biy+Cy'+Dx + Ey+F. 



PAETIAL DIFFERENTIATION 197 



n / o . V -» . ov- . du 2anzu 



du _^ 2 6^Jiyu 



4. u = arc sin - • -4n«. — = 



cu X 



c'y y VyS — X* 

5. u = x*. -4?w. — = yzff-^ ; 

ax 

au 

— = xJ'logx. 
oy 

6. u = axV« + tey'«* + cy« + dxz». ilrw. — = 3 ax'^^z + fry*** + d2* ; 

dx 

— = 2ax*yz + 3tey«2:* + 6q^; 

ay 
a^ 

7. u = x^ — 2 xy* + 3 x^y* ; show that x \- y — = 6tt. 

dx (iy 

8. u = — ^— ; show that x h y — = w. 

X + y ax ay 

9. u = (y - z) (2 — x) (x — y) ; show that — + — H = 0. 

as dy dz 

10. u = log (e* + c"') ; show that h — = 1. 

dx dy 

11. u = ; show that 1 = (x + y — l)u. 

e* + CI' dx dy 

du du 

12. u = xi'y* ; show that x H y — = (x + y + log u) u. 

dx dy 

13. u = log(x« + y» + «»-3xyz); show that — + ^'^ + — = ^ 



dx dy dz x-\-y-\-z 

14. tt = c*8in y + c'' sin x ; show that 

/?? y + ( — y= «'* + c't' + 2 e'+vsin (x + y). 

15. tt = log (tan X + tan y + tan z) ; show that 

sin 2 X 1- sin 2 y 1- sin 2 z — = 2. 

dx dy dz 

16. Let y be the altitude of a right circular cone and x the radius of its base. 
Show (a) that if the base remains constant, the volume changes ^irx^ times as fast 
as the altitude ; (b) that if the altitude remains constant, the volume changes f rxy 
times as fast as the radius of the base. 



198 DIFFERENTIAL CALCULUS 

17. A point moves on the elliptic paraboloid z = — f- — and also in a plane par- 

9 4 

allel to the XOZ plane. When x = 3 f t. and }» increasing at the rate of 9 It. per 
second, find (a) the time rate of change of z ; (b) the magnitude of the velocity of 
the point ; (c) the direction of its motion. 

Ana. (a) r, = 6 f t per sec. ; (b) r = 3 Vis ft. per sec. ; 

(c) r = arc tan f , the angle made with the XOT plane. 

18. If, on the surface of Ex. 17, the point moves in a plane parallel to the plane 
YOZ, find, when y — 2 and increases at the rate of 5 ft. per sec, (a) the time rate 
of change of z ; (b) the magnitude of the velocity of the point ; (c) the direction of 
its motion. ^^ ^^^ ^ ^^ ^^ ^^ ; (b) 6 V2 ft. per sec. ; 

(c) r = — ) the angle made with the plane XOT, 
4. 

136. Total derivatives. We have already, on page 57, considered 
the differentiation of a function of one function of a single inde- 
pendent variable. Thus, if 

y =f{v) and v = <^(a;), 

it was shown that 

dy ^dy dv 

dx dv dx 

We shall next consider a function of two variables, both of which 
depend on a single independent variable. Consider the function 

u=/(x, y), 

where x and y are functions of a third variable t. 

Let t take on the increment Af, and let A:r, Ay, At^ be the corre- 
sponding increments of x^ y, u respectively. Then the quantity 

Aw =f{x -h Aa:, y 4. Ay) -f(x, y) 

is called the total increment of u. 

Adding and subtracting /(a:, y 4- Ay) in the second member, 

{A) Lu = [fix 4- At, y -h Ay) -/(:r, y 4. Ay)] 4- [f{x, y 4- Ay) -/(x, y)] . 

Applying the Theorem of Mean Value, (44), p. 168, to each of 
the two differences on the right-hand side of (^), we get, for the 
first difference, 

(5) /(a? 4- A2:, y + Ay) -/(a:, y 4- Ay) =/,7a; 4- 5i • A2;, y 4- Ay) Aar. 

ras X, Aas Aor, and since x varies while .v + Ay remainsl « 
Lconstant, ve get the partial derirative with respect to x.\ 



PARTIAL DIFFERENTIATION 199 

For the second difference we get 

( 0) /{x, y + Ay) -/(x, y) = //(x, y + ^, • Ay) Ay. 

[as y, Aa aa Ay, and eince y yarles while x remaine oon-l 
Btant, we get the partial derivative with respect to y.J 

Substituting (B) and ((7) in (A) gives 
(p) Au = f^(x 4- ^j . Aa;, y 4- Ay) Lx + //(z, y -f- ^2 • Ay) Ay, 
where 0^ and 5, are positive proper fractions. Dividing (JP) by A^ 

(E) ^ =/;(ar + ^, . Aa:, y + Ay) ^ +/;(x, y + ^, . Ay) ^ . 
Now let A^ approach zero as a limit, then 



r Since Aj; and Ay converge to xero with A<, we get "1 

a/IL*o/-'<* + ^ • ^. y + Ay) -/-'(ar, y), and l*/^**o/»'('» ^ + *« ' Ay)-/,'(x. y), 



a/IL*o/-'(* + ^ • Aor, y + Ay) -/.'(ar, y), and ^*™J* 
L. /vX^i y) And /y'(x, y) being asaumed continuouB 

Replacing /(ic, y) by u in (i^), we get the total derivative 

^ ' <l« " aa; de "*" c)y dt ' 

In the same way, if ^, . 

•^ w =/(a;, y, 2), 

and 2;, y, 2 are all functions of ^, we get 

/CA^ — — — — 4- — ^ -A- — — 

^ ^ dt" dx dt'^ dy dt'^ dz dt* 

and so on for any number of variables.* 

In (49) we may suppose t = x; then y is a function of x^ and u is 
really a function of the one variable 2:, giving 

/«\ dUf _ du du dy 

dx "doc dy doD * 

In the same way from (50) we have 

dx" dx dy dx dz dx* 

* This Is really only a special case of a general theorem which may be stated as follows : 
If u is a function of the independent variables x, y, 2, • • •, each of these in turn being a funo* 
tfon of the independent variables r, «, f, • . ., then (with certain assumptions as to continuity) 

c^ii du dx du dy du dz 
dr dx dr dy dr dz dr 

du du 
and similar expressioitf hold for , — , etc. 



200 DIFFERENTIAL CALCULUS 

s 

The student should observe that — and -r- have quite different 

dx ax 

dti 
meanings. The partial derivative — is formed on the supposition 

ox 

that the particular variable x aloiie varies^ while 

du^ limit /Aw\ 
dx Ax=0\AxJ' 

where Au is the total increment of u caused by changes in all the 
variables, these increments being due to the change A2; in the inde- 
pendent variable. In contradistinction to partial derivatives, — , — 
are called total derivatives with respect to t and x respectively.* 

Ex. 1. Given m = sin -» x = c*, y = <* ; find — . 

y dt 

^ . .. du- I X du X X dx , dy .. 

SolxUum. — = - cos - » — = cos - ; -— = c*, -=^ = 2f. 

dx y y dy y^ y dt dt 

du 6^ 6^ 

Substituting in (49), — = (t — 2) — cos — • Ans. 

Ex. 2. Given « = ««(y - *), y = asinx, r = cosx; find *?. 

dx 

du . du du dy dz 

Solution. — = ae^ (y - 2), — = e«*, — = — c*^ ; -=^ = a cos x, — = — sin x. 

dx dy dz dx dx 

Substituting in (62), 

du 

-— = a€f'{y — z) + a€f^coaX'\- e^Binx = e**{a^ + l)sinx. Ans, 

dx 

Note. In examples like the above, u could, by substitution, be found explicitly 
in terms of the independent variable and then differentiated directly, but generally 
this process would be longer and in many cases could not be used at all. 

137. Total differentials. Multiplying (49) and (50) through by dt, 
we get, (§ 104, p. 144), 

(58) ^^ "= ^ '^ + ay ^^' 

(54) au^ — dx + -dy + — dzi 

o It should be observed that x- has a perfectly definite value for any point (x, y)t vhile — 

ex dx 

depends not only on the point (x, y) but also on the particular direction chosen to reach that 

point. Hence ^^ 

V- is called a point function ; while 

dx 

— is not called a point function unless It is agreed to approach the 
dx 
point from some particular direction. 



PARTIAL DIFFERENTIATION 201 

and so on.* Equations (53) and (54) define the quantity du^ which 
is called a total differential of u or 3, complete differential^ 

and -r-"^? :r^!/j tt"^ 

dx dy ^ dz 

are called partial differentials. These partial differentials are 
sometimes denoted by d^u^ d^Uy d^u, so that (54) is also written 

du = d^u 4- d^u 4- rf,w. 

y 
Ex. 1. Given u = arc tan - : find du, 

X* 

'Solution, 



Substituting in (53), 



tu 


^ 


y 


bu 
cy 


X 


dx 


{B» + y« 




du 


_^xdy — 


ydx 


Ana. 



x2 + ya 



Ex. 2. The base and altitude of a rectangle are 6 and 4 inches respectively. 
At a certain instant they are increasing continuously at the rate of 2 inches and 
1 inch per second respectively. At what rate is the area of the rectangle increasing 
at that instant ? 

Solution. Let x = base, y = altitude : then u = xy = area, — = y, — = x. 

dx dy 

Substituting in (40), 

, A\ du dx , dy 

<^> ^=''^+^*- 

But X = 6 in., y = 4 in., — = 2 in. per sec, - - = 1 in. per sec. 

dt dt ' 

du 
,'. — = (8 + 5) sq. in. per sec. = 18 sq. in. per sec. Ans, 

KoTB. Considering du as an infinitesimal increment of area due to the infinites- 
imal increments dx and dy^ du is evidently the sum of two thin strips added on to 
the two sides. For, in du = ydx + xdy (multiplying (A) 



■^Ji^y^^^y.^^:^. 



^ w 

^ 



' ydx = area of vertical strip, and ^ 

xdy = area of horizontal strip. 

But the total increment Au due to the increments dx and dy ^ 
^ Att = ydx + xdy + dxdy, 

Henoe the small rectangle in the upper right-hand comer (— dxdy) 5a 

is evidently the diif erence between Au and du. This figure illus- 
trates the fact that the total Increment and the total differential of a function of several vari* 
ables are not In general equal (see p. 141). 

* A geometric interpretation of this result will be given on p. 274. 



r 



202 DIFFERENTIAL CALCULUS 

138. Differentiation of implicit functions. The equation 

(A) fix, y) = 

defines either 2; or y as an implicit function of the other.* It rep- 
resents any equation containing x and y when all its terms have 
been transposed to the first member. Let 

{B) u^f{x,y); 

du 
But from (A), f{x, y) =0. .-. w = and — = ; that is, 

^ ' dx dy dx 

Solving for -^» we get 

dji^ du dy 

dy 

a formula for differentiating implicit functions. This formula in 
the form {C) is equivalent to the process employed in § 75, pp, 83, 
84, for differentiating implicit functions, and all the examples on 
p. 85 may be solved using formula (55). Since n 

(D) /(^y)=o 

for all admissible values of x and y, we may say that (55) gives the 
relative time rates of change ofx and y which keep f(x, y)from chang- 
ing at all. Geometrically this means that the point {x, y) must 
move on the curve whose equation is (i>), and (55) determines the 
direction of its motion at any instant. Since 

we may write (55) in the form 

* We assume thai a small change in the yalue of x causes only a small change in the yalue of y 

t It is assumed that -- and -- exist. 

ex dy 



PARTIAL DIFFERENTIATION 



203 



Ex. 1. 6iyenxV + 8in]^ = 0; find-r^. 

dx 

SolvUon, Let/(x, y) = x^ + any, 

^ = 2ajy*, ^ = 4xV + co8y. . .froin(66a),^ = - , , ^^ 

dx 'ay ^"dx 4xV + co8y 



^n«. 



Ex. 2. If X increases at the rate of 2 inches per second as it passes through the 
value X = 3 inches, at what rate roust y change when y = 1 inch, in order that the 
function 2xy> — 8xV shall remain constant ? 

Solution. Let /(x, y) = 2 xy> - 8 x^ ; then 

^ = 2y« - 6xy, ^z^^xy-- Ssfi. 
dx dy 



Suh8titutingm(66a), 

dy _ 2y«~6xy 
dx 4xy — 8x^ 



»or, 



dy 

dif_ Zy^-exy 



dx 
d< 



4xy - 3x« 



By (A), p. 164 



(56) 



ft/f dii 

But X = 8, y = 1, -- = 2. .*. ^ = — 2,^ ft. per second. Ans, 

dt dt 

Let F be the point (2;, y, 2) on the surface given by the equation 

and let PC and AP be sections made by planes through P parallel 
to the YOZ and XOZ planes respectively. Along the curve -4P, 
y is constant, therefore from {JE)^ z is 
an implicit function of x alone, and we 
have from (55 a) 

to"" aip' 

giving the slope at P of the curve -4P, 
§ 183, p. 193. 

— is used instead of -:;- in the first member since z was originally 
dx dx 

from {II) an implicit function of x and y, but (56) is deduced on 
the hypothesis that y remains constant. 
Similarly the slope at P of the curve PC is 

dF 

(57) ^ = -iy. 

^ *' by dF 

dz 




204 DIFFERENTIAL CALCULUS 



EXAMPLES 

Find the total derivatiyes, using (49), (50), or (61) in the following six examples. 

du 

1. u = z^ + y^ + zy,z = &inz,y = e*. Ans, -'=3c»*+c»(8inx+cosx)+8in2z. 

ax 

du c*(l +x) 

2. tt = arc tan (xy). y = €f, Ans. — - = — -^^ — — — • 

3. tt = log(a2-p2), p = (,8in^. Ans. — = -2tan^. 

4. u = iJ* + cy, « = log », y = 6^. An», ■— - = h tw*. 

as 9 

du « 

5. u = arc sin (r — «), r = 3 f, » = 4 1". -An«. — = 



dt Vl-<i 

e«f(y — 2) . ^ <*« ^, • 

6. u = — ^^^ ^t y = asinx, z = cosx. Ans, --- = e«^smx. 

a* + 1 dx 

Using (53) or (54), find the total differentials in the next eight examples. 

7. tt = 6y^+ cx« + yy* + «c. Ans. du = (by^-^2cx+e)dx-\'(2byx+Sgy^dy. 

8. u = log XV. Ans. dti = - dx + log zdy. 

X 

sin X 

9. us5«^*. iliM. du = j^* log y COS xdxH dy. 

10. u = x^K Ans. du = tt(!^dx + J^^dyV 

a-^-t A A 2(«tt-«d«) 

11. u = . Ans. du = — ^^ — ^ . 

« - t (« - 0* 

12. u = sin (i)g). Ans, du = cos (pg) [gdp + pdq], 

13. u = xi^. Ans, du = xff-^ {yzdx + zx log xdy + xy log xdz). 

14. u = tan* ^ tana ^tan«f. Ans. du = 4uC-^ + -^ + -^V 

\sin20 sin2« sin2^/ 

15. Assuming the characterlstle equation of a perfect gas to be 

vp = i?t, 
where v = volume, p = pressure, t = absolute temperature, and R a constant, what 
is the relation between the differentials do, dp, dt ? Ans. vdp + pdv = Rdt. 

16. Using the result in the last example as applied to air, suppose that in a 
given case we have found by actual experiment that 

t = 300® C, p = 2000 lbs. per sq. ft, c = 14.4 cubic feet. 

Find the change in p, assuming it to be uniform, when t changes to SOP C, and 
to 14.5 cubic feet. 22 = 90. Ans. - 7.22 lbs. per sq. ft. 



PARTIAL DIFFERENTIATION 206 

In the remaining examples find -^ i using formula (66 a). 

ox 

^ *^ ' ^ ' dx y 2 (x« + ys) + o2 

18. e»-e» + xy = 0. -4im. ~ = ^""^ « 

^ *> . / V « A ^ ^y y [cos (xy) — c«y — 2 x] 

19. sm(xy)-c*i'-x«y = 0. -4)w. -^ = iLL — ^^J'^ i. 

•dx x[x + e'»'-co8(xy)] 

20. sin X sin y + cosx cos y — y = 0. 



139. Successive partial derivatives. 

then, in general, 

du J du 
— and — 

ex dy 

are functions of both x and y, and may be differentiated again 
with respect to either independent variable, giving successive par- 
tial derivatives. Regarding x alone as varying, we denote the 
results by 

a^ a^ as^ ru 

dx"' d2^' dx*' '"' aaf' 
or, when y alone varies, 

^ ^ d^ dTu 

a/' a/' ay*' '"' ay' 

the notation being similar to that employed for functions of a 
single variable. 

If we differentiate u with respect to a;, regarding y as constant, 
and then this result with respect to y, regarding x as constant, we 
obtain 

— f — b which we denote by • 

dt/\dxj ^ dydx 

Similarly, if we differentiate twice with respect to x and then 
once with respect to y, the result is denoted by the symbol 

av 

dydJi?' 



206 DIFFERENTIAL CALCULUS 

140. Order of differentiation immaterial. Consider the function 
f(x^ y). Changing x into x + Aa; and keeping y constant, we get 
from the Theorem of Mean Value, (44), p. 168, 

(A) f{x + l!ix,y)^f{x,y)^£iX'fJ{x + e^^x,y). 0<^<1 

[ass x» Aass Ax, and since x Taries while y remaina oon-l 
Btant, ^e get the partial derivatiTe with respect to x.\ 

If we now change y to y + Ay and keep x and Aa; constant, the 
total increment of the left-hand member of {A) is 

{B) [f{x + A;r, y + Ay) -f{x, y + Ay)] - [f{x + Aa:, y) -f{x, y)]. 

The total increment of the right-hand member of {A) found by 
the Theorem of Mean Value, (44), p. 168, is 

{O) AxfJ{x + ^Ax, y ^ Ay) - AxfJ {x -^ 0' Ax, y). < ^^ < 1 

= AyAxfJ'(x + d,^Ax,y + d^.Ay). < ^, < 1 

[ae y, Aas Ajft and since y varies while x and Ax remainl 
constant, we get the partial deriratiTo with respect to y. J 

Since the increments {B) and (C) must be equal, 

(2>) [f{x + Aa:, y + Ay) -f(x, y + Ay)] - [f(x + Ax, y) ^f(x, y)] 

= AyAxf^^x + e^.Ax, y + ^,. Ay). 

In the same manner, if we take the increments in the reverse order, 

{E) [f{x + Aa:, y + Ay) -/(a: + Aa:, y)] ~ [f(x, y + Ay) -f{x, y)] 

= AxAyf^\x + ^, . Aa;, y + ^, . Ay), 

0^ and 0^ also lying between zero and unity. 

The left-hand members of (2>) and {E) being identical, we have 

{F) /^> + ^i-Aa:,y + ^,.Ay)=/^"(a: + ^..Aa:,y + ^,.Ay). 

Taking the limit of both sides as Aa; and Ay approach zero as 
limits, we have * 

since these functions are assumed continuous. Placing 

{(jt) may be written 

(58) ^'^ ^ 



dydx dxdy 

* Assuming the continuity of the first partial derivatlTee and the existence and continuity of 
^."and/,.". 



PARTIAL DIFFERENTIATION 207 

That is, the operations of differewtiating with respect to x and with 
respect to y are commutative. 

This may be easily extended to higher derivatives. For instance, 
since (58) is true, 

da^dy "" dz \dxdy ) " dxdydx dxdy \ dx) dydx \ dx) "" dydj? ' 
Similarly for functions of three or more variables. 



Ex. 1. Given u = x»y — Sx^y* ; verify 



dydx dxdy 



SolutioTi. — = 8*«y-6xy«, — = 3x2_i8zy2, 
dx dydx 

— = x» - 9xV» -^ = 3x« - 18xy«; hence verified. 
dy dxdy 



1. u = cos(x + 2^); 

2. u = ?^; 

3. tt = ylog(l + 3cy); 

4. tt = arc tan - ; 
6. u = Bin(^^); 

6. u = 6c«y«« + 3c»x»z«+ 2c«x*y - xyz; show that ^J' =12(e'y + ci'z + ex) 

dx'^cydz 

7. tt = e*i" ; show that = (1 + Sxyz + xV**)"- 

axayaz 

8. u = — ^; showthatx— - + y-— - = 2 — . 

x + y 5x* dxay ax 

9. tt = (x« + y*)*; 8howthat8x---- + 3y---+ — = 0. 

^ ' axey dy^ dy 

* V i ^u - '- ' 

10. M = y«z%* + z^V + xVc*; show that ^— — — - = e« + e* + A 

cx^cy-'cz'* 

1 c*u dhi dhi 

11. u = (x« + y» + «r*;«howthat — + — + — = 0. 



veriiy 


dydx 


dxdy 


verify 


dhi 
dydx 


dhi 
dxdy 


verify 


dhi 
dydx 


dhi 
dxdy 


verify 


dhi 
df^da 


dhi 
'd8cr^'* 


verify 


dhi 
dedip^ 


a»tt 
dipf^de 


T.1tT • a1 


linw thfl. 


, dhi 




CHAPTER XVIII 
ENVELOPES 

• 

in. Family ot curves. Variable parameter. The equation of a 
curve generally involves, besides the variables x and y, certain con- 
stants upon which the size, shape, and position of that particular 
curve depend. For example, the locus of the equation 

(A) ix-af + y^^r' 

is a circle whose center lies on the axis of X at a distance of a from 

the origin, its size depending on the 

radius r. Suppose a to take on a 

series of values, then we shall have a 

corresponding series of circles differing 

in their distances from the origin, as 

shown in the figure. 

Any system of curves formed in this way is called a family of 

curves^ and the quantity a, which is constant for any one curve, 

but changes in passing from one curve to another, is called a vari- 

able parameter. 

As will appear later on, problems occur which involve two or 
more parameters. The above series of circles is said to be 9, family 
depending an one parameter. To indicate that a enters as a vari- 
able parameter it is usual to insert it in the functional symbol, 

*"^= /(a:,y,a)=0. 

142. Envelope of a family of curves depending on one parameter. 

Any two neighboring curves of a family will in general intersect.* 
If the corresponding values of the parameter are a and a + Ao, the 
point or points of intersection, if these exist, will in general tend 
to definite limiting positions (points) as Aa approaches zero. The 
locus of all such limiting points is called the envelope of the family 

* An exception to this would be the Bystem of concentric circles we get from {A) when a is 
constant and r varies, no two of which would intersect. 

208 




ENVELOPES 209 

of curves. Thus, in the last figure, the limiting positions of the 
points of intersection of the circles are all on the straight lines AB 
and CD, which form therefore the envelope of the family of cii'cles. 

Now find the equation of the envelope of a family 
of curves depending on one parameter. Let 

(B) /(x, y, a)=0 and 

(0) f(x,ff,a + Aa)=0 

be two neighboring curves of the same family ^ '(b) 
intersecting at a point (2/, y') ; and let us find the limiting position 
of this point of intersection as Aa approaches the limit zero. 

We can find the equation of a third curve through (a/, y') by 
applying the Theorem of Mean Value, (44), p. 168, to (B) and ((7), 
regarding a as the variable and x and y as constants. For we have 

(D) f{x,y,a-\-Aa)-f{x,y,a) = AafJ(x,t/,a-{-0.Aa). 0<^<1 

Since P' lies on both of the curves (B) and ((7), the left-hand 
members of their equations vanish for x = xf and y = y'. Hence 
the left-hand member of (2>) must vanish for the same values, and 
consequently the right-hand member also. Therefore 

(J?) fJ(x,y,a + 0.Aa)^O 

is the equation of a third curve passing through the intersection 
of (B) and ((7). If then (B) and (O) intersect in a point which 
approaches a fixed point as a limit as Aa approaches zero, we get 
in general 

(F) /;(2^y,a)=0 

as the equation of a curve which passes through the limit of the 
intersection of (B) and (C), In general (F) is distinct from (B) 
and therefore has a definite intersection with it. 

Since the coordinates of the points on the envelope satisfy both 
(F) and (jB), its equation is found by eliminating a between these 
equations. The equation of the envelope is therefore a new rela- 
tion between x and y that is independent of a.* 

* By definition we should solve (B) and (C) Bimnltaneously for their point of intersection and 
then pass to the limit. In practice, however, it is found to he more convenient first to pass to 
the limit and then solve for x and y, just as we do here. It is not self-evident hy any means that 
these two processes give the same results in all cases, hut it is a fact that the results are iden- 
tical in all the applications made in this hook. 



210 



DIFFERENTIAL CALCULUS 



On account of the*process of elimination that is involved, no 
detailed method of procedure can be given for finding the envel- 
ope that will apply in all cases. In a large number of problems, 
however, the student may be guided by the following 

General directions for finding the envelope. 

First step. . Differentiate with respect to the variable parameter ^ con- 
siderinff all other quantities involved in the given equation as constants. 

Second step. Solve the resvltfor the variable parameter. 

Third step. Svhstitute this value of the variable parameter in the 
given equation. This gives the equation of the envelope. 

Ex. 1. Find the envelope of the straight line y = mx + — » where the slope m is 
the variable parameter. 



Solviion, 



First step. 






o = x — 



m- 



2 



Second step. m = ± 




Third step, y = ± \/| « db -^1 P = ± 2 V^, 




and squaring, y^ = 4|>x, a parabola, is the equation of 
the envelope. The family of straight lines formed by 
varying the slope m is shown in the figure, each line being tangent to the envelope, 

for we know from Analytic Geometry that i^ = mx + — is the tangent to the parab- 
ola y' = 4px expressed in terms of its own slope m. 

143. The envelope touches each curve of the family at the limiting 
points on that curve. 

Q-eometrical proof Let A^ B^ C hQ three neighboring curves 
of the family, A and B ii^tersecting at P, and B and C at Q. 
Draw TT' through P and Q. Now let A and C approach coin- 
cidence with B^ that is, let A^ B^ C become consecutive curves * 
of the family. Then TT^ becomes a tangent to B^ having two 



*The limiting position of any point of intersection (as P in figure, p. 211) is sometimes called 
the point of intersection of two e(m8ee%U%ve curves of the family. Similarly the line which TT* 
approaches as P approaches Q, i.e. the tangent to iB at Q, is said to pass through two c<m»ee%Uive 
points of the curve. Of course there is no one curve that is consecutive to another nor any one 
point that is consecutive to another in the ordinary sense of the word, but geometrical considera- 
tions have suggested the above phraseolo^v nnd It is understood to be merely a brief way of 
indicating the actual condition of affairs as stated in the definitions. 




ENVELOPES 211 

consecutive points P and Q in common with it. But then P and 
Q will also become consecutive points of 
the envelope by definition ; hence TT' will 
at the same time become a tangent to the 
envelope. Therefore B and the envelope have a common tangent ; 
similarly for every curve of the family. Thus in the example of 
the last section we noticed that the parabola had the family of 
straight lines as tangents. 

Analytical proof. Consider the family of curves represented by 

(A) f(x, y, «) = 0. 

dy 
The 8loi>e at any point on (il) is the value of -^ from 

dx 

(B) ?^ + ^J? = 0; (61), p. 190 

dx dy dx 

where, in differentiating, a must be kept constant. 

From the previous section we know that the envelope of the family of curves is 
found by eliminating a between {A) and 

(C) |-/(x, y, a) = 0. 

da 

If we suppose (C) solved for a in terms of z and y and the result substituted 
in (A), it is evident that equation (A) would then be the equation of the envelope. 
Hence the slope of the envelope may be found by taking the total derivative of (^), 
(52), p. 199, regarding a as a certain function of x and y determined by (C). This gives 

dx dy dx da dx 

Suppose now that the coordinates of the point (x, y) satisfy both (A) and (C) ; 
that point is therefore on the curve (A) and also on the envelope ; and, by (C), the 
last term in (D) vanishes, reducing (D) to the same form as {B). Hence at the 
point (x, y) the slope is the same for the curve {A) and the envelope, so that a 
limiting point of intersection on any member of the family is a point of contact of 
this curve with the envelope.* 

144. Parametric equations of the envelope of a family depending 
on one parameter. Instead of finding the equation of the envelope 
in rectangular form by the method of § 142, p. 210, it is sometimes 
more convenient to get the equations of the envelope in para- 
metric form by solving 

f(x, y, a) = and —f{x, y, a) = 
for X and y in terms of a. Thus : 

• In the special caae when ^ = ^ = or -• =0 for all points of oar locus this reasonlr 

ex cy cy 



212 



DIFFERENTIAL CALCULUS 



Ex. 1. Find the envelope of the familyiof straight lines x cos a + y sin a = p, a 
being the variable parameter. 

Solution, (A) z cos a + y sin a = p. 

Differentiating (A) with respect to a, 

(B) — X sin a + y cos a = 0. 

Multiplying {A) by cos a and {E) by sin a and sub- 
tracting, we get , = j,co8.. 

Similarly, eliminating x between {A) and (B), we 

^ 2^ = p sin a. 

The parametric equations of the envelope are there- 
fore 

(C) 




Iz=p cos a, 
y = psino; 

a being the parameter. Squaring equations (C) and adding, we get 

x« + y« = p«, 

the rectangular equation of the envelope, which is a circle. 




Ex. 2. Find the envelope of a line of constant length a, whose extremities move 
along two fixed rectangular axes. 

Solution. Let AB = a in length, and let 

{A) X cos a + ysino— p = 

be its equation. Now as AB moves always touching 
the two axes, both a and p will vary. But p may be 
found in terms of a. For, ^O = ^ B cos a = a cos a, 
and p = ^ sin a = a sin a cos a. Substituting in (A\ 

(B) X cos a + y sin a — a sin a cos a = 0, 
where a is the variable parameter. Differentiating (B) with respect to a, 

(C) — X sin o -f y cos o + a sin* o — a cos* o = 0. 

Solving (B) and (C) for x and y in terms of a, we get 

IX = a sin' a, 
y = a cos' a, 

the parametric equations of the envelope, a hypocycloid. 

The corresponding rectangular equation is found from equations (D) by eliminat 
ing a as follows : 

x' = a' sin* a. 

yi = a* cos* a. 
Adding, x' + y' = a*, 

the rectangular equation of the hypocycloid. 



ENVELOPES 218 

146. The evolute of a giveo curve considered as the eovelope of 
its normals. Since the normals to a carve are all tangent to the 
evolute, §129, p. 186, it is evident that tU 
evolute of a carve may also be defined as the 
envelope of iti normals ; that is, as the locua 
of the ultimate intersections of neighboring 
normals. It is also interesting to notice 
that if we find the parametric equations of 
the envelope by the method of the previoua 
section, we get the coordinates x and y of 
the center of curvatnre ; bo that we have 
here a aecond method for finding the cod'rdtnates of the center of 
curvature. If we then eliminate the variable parameter, we have 
a relation between x and y which ia the rectangular equation of 
the evolute (envelope of the normals). 



Solution. The equation of the normal at an; point (z', ^ is 

'-'■ — Ti^"^ 

from (2), p. 90. Aa we are considering the normals all along the curre, both 
z' and y' will vary. Eliminating x' by means of v^ = 4 px', we get the equation 
of the normal to be 

Considering j/' as the yariable parameter, we wish to find the envelope of Uiis 
family of normals. DlBerenti sting (A) with respect to v', 

_1 = ^_^ 
Spi 2p 
and solving for x, 

Subetitating this value of x in {A) and solving for v, 

y^-yi. 

" 4p' 

(B) sud (C) are then the coordinates of the center of curvature of the parabola. 
Taken together, (S) and (C) are the parametric equations of the evolute in terms of 
the paramet«T y". Eliminating y" between (if) and (C) gives 

27py* = 4(i-2p)", 
the rectangular equation of the evolute of the parabola. This is the same result we 
obtained in Ei. 1, p. 183, by the first method. 



(C) 



214 DIFFERENTIAL CALCULUS 

146. Two parameters connected by one equation of condition. 
Many problems occur where it is convenient to use two parameters 
connected by an equation of condition. For instance, the example 
given in the last section involves the two parameters J and y' 
which are connected hy the equation of the curve. In this case 
we eliminated x\ leaving only the one parameter y'. 

However, when the elimination is difficult to perform, both the 
given equation and the equation of condition between the two 
parameters may be differentiated with respect to one of the param- 
eters, regarding either parameter as a function of the other. By 
studying the solution of the following problem the process will be 
made clear. 

s coincide und 



SobiXvi-n. (A) i- + ^ = 1 

la the equRtion of the elllpM where a and b are the variable panuneters connected by 
the equation 

(B) iroft = k, 

tab being the area of an ellipse whose 
semiaxes are a and h. Differentiating 
(A.) and (B), regarding a and b Eis vari- 
ables and X and y ae constants, we have, 
naing diSereiitjala, 

—J- + ^ = 0. from {A), 
and Ma + adb = 0, from (B). 

Transposing one tenn in each to tbe 
second member and dividing, we get 



Therefore, from {A\, —-- and — = - , 

giving a = ± I V^ and h = ±y^. 

Substituilng these values in (B), we get the envelope 

a pair of conjugate rectangular hyperlwlas (see flgora). 



ENVELOPES 215 



1. Find the envelope of the family of straight lines y = 2 mx + m*, m being the 
Tariable parameter. Ana. x = — 2 m«, y = — 3 m* ; or, 16 y« + 27 z* = 0.» 

2. Find the envelope of the family of parabolas ^ = a(x — a), a being the 
variable parameter. An8. x = 2 a, y = ± a ; or, y = i J x. 

3. Find the envelope of the family of circles x^ + (y — /S)^ = r^, /3 being the 
variable parameter. Ana, x=:±r. 

4. Find the equation of the curve having as tangents the family of straight lines 
y =:mx± Va*wM-^i the slope m being the variable parameter. 

Ana. The ellipse Wx^ + aV ^ o?t^' 

6. Find the envelope of the family of circles whose diameters are double ordi- 
nates of the parabola y^ = 4px. Ana. The parabola y^ = 4p (p + x). 

6. Find the envelope of the family of circles whose diameters are donble ordi- 

nates of the ellipse V^^ + a^r/^ = a^fta. ^ „^ „. x« y^ ^ 

Ana. The ellipse — — -- + ^ = 1. 

7. A circle moves with its center on the parabola y^ = 4 ox, and its circumference 

passes through the vertex of the parabola. Find the equation of the locus of the 

points of ultimate intersection of the circles. 

Ana. The cissoid y^ (x + 2 a) + x* = 0. 



8. Find the curve whose tangents are yz=lx± VaP -|_ ^ + c, the slope I being 
supposed to vary. Ana. i (ay^ + 6xy + cx^) = 4ac -h^. 

9. Find the evolute of the ellipse ¥x'* + a^> = a^b^, taking the equation of 
normal in the form &y = ax tan - (a«-62)8in 0, 

the eccentric angle </> being the parameter. 

-47W. X = — — — COS* 0, y = — -■ — sin* 0; or, (ox)' + (&y)' = (o^— ft^jl. 
a 

10. Find the evolute of the hypocycloid x' + y' = a*, the equation of whose 
normal is y cos t - x sin t = a cos 2 r, 

T being the parameter. Ana. (x + y)' + (x - y)' = 2 a'. 

11. Find the envelope of the circles which pass through the origin and have 
their centers on the hyperbola x^ — y^ = ^a. 

Ana. The lemniscate (x^ + y^)^ = a^ (x^ - y«). 

12. Find the envelope of a line such that the sum of its intercepts on the axes 
equals c. Ana. The parabola x* + y* = c*. 

* When two answers are given, the first Is in parametric form and the second In rectangnhir 
form. 



216 



DIFFERENTIAL CALCULUS 



Finl the CBvdope of tke fualj of 



of iiB 



14. Find the CBTeiope of 
becwmn the 

cqoal to L 

15. ftojedilei are fired froMm 



the Sam 
hTpocjdoid x^ + y* = A 



of the 



A9f^ Aeq 




gun vishan 
e»begi 
in the 
of an poBible 



anr 



1 .»» 



trajcctoriea. the 



and oidi that the 
is eonstant and 
adea are (x :^ jr)* = I*. 

the gun 

is kefit always 

is the enielope 

of the 



»=*' 



«r,« 






CHAPTER XIX 

SERIES 

147. Introduction. A series is a succession of separate numbers 
which is formed according to some rule or law. Each niunber is 
called a term of the series. Thus, 

is a series whose law of formation is that each term after the first 
is found by multiplying the preceding term by 2 ; hence we may 
write down as many more terms of the series as we please, and 
any particular term of the series may be found by substituting the 
number of that term in the series for n in the expression 2*"*, which 
is called the general or nth term of the series. 



In the following six series : 






(a) Discover by insi)ection the law of formation ; 




(b) write down several terms more 


in each; 




(c) find the nth or genercU term. 






Series 




nth term 


1. 1,3,9,27,.... 




s^-i. 


2. — a, + a*, — a», +«*,•• •• 




(- a)". 


3. 1, 4, 9, 16, .... 




n2. 


. X* X« X* 

*• ^' 2' 3^ 4^-- 




X" 




n 


B. 4, -2, +1, -i, .... 




.4(-i).-«. 


« 3y 5ya 7^3 
2 6 10 




n»+l *^ 



Write down the first four terms of each series whose nth or general term is given 
below. 

nth term Series 

7. fA£\ X, 4x2, 9x», 16x*. 

X" X x^ x» X* 



8 



l+Vn ^^^^^ .^^^^^ I + V2 I+V3 1+v^ 



XT 




218 DIFFERENTIAL CALCULUS 



rUhterm 


iSeriea 


Q n-f2 


3 4 6 6 
2' 9* 28' 65 


10. ^. 

2» 


12 3 4 

2' 4' r 16- 


^^ (l0g0)»5B» 


loga-x log^tt'X* log^a-x* log* a* a:* 
1 • 2 ' 6 • 24 • 


|2n-l 





148. Infinite series. Consider the series of n terms 

(A) 1 ^ ^ ^ ^ • 

W 1. 2' 4' 8 ■■■' 2^*' 

and let S, denote the sum of the series. Then 
(5) s, = l + l + l + l + ... + ±,. 

Evidently S^ is a function of w, for 
when n = 1, ^1 = 1 =1, 

when 71 = 2, S^ = l + \ =1J, 

when M = 3, ^3 = 1 + 14-1 =:1|, 

when n = 4, ^^ = 1+1 + 1 + 1 =:1J, 



when n = n^ ^. = 1H-x + t + 7^H h :r— r = 2 — -r— r-* 

Mark off points on a straight line whose distances from a fixed , 
point correspond to these different sums. It is seen that the 

point corresponding to any sum bisects the distance between the 
preceding point and 2. Hence it appears geometrically that when 
n increases without limit 

limit /S. = 2. 

* Fonnd by 6, p. 1, for the sum of a geometric series. 



SERIES 219 

We also see that this is so from arithmetical considerations, for 



limit^ Umit/2_ 1 \ 2. 



I since when n inoreases without limit, ^»proaohes xero as a limit.] 

We have so far discussed only a particular series (A) when the 
number of terms increases without limit. Let us now consider the 
general problem, using the series 

(C) Mj, u„ u„ u^y ..., 

whose terms may be either positive or negative. Denoting by S^ 
the sum of the first n terms, we have 

/Sf, = Wi + t^j, 4- w, + . . . + w., 

and ^S; is a function of n. If we now let the number of terms (= n) 
increase without limit, one of two things may happen : either 

Case I. 8^ approaches a limit, say it, indicated by 

Case II. S^ approaches no limit. 

In either case {O) is called an infinite series. In Case I the infinite 

series is said to be convergent and to converge to the value ti^ ot to 

have the value u^ or to have the sum u. The infinite geometric 

series discussed at the beginning of this section is an example of 

a convergent series, and it converges to the value 2. In fact, the 

simplest example of a convergent series is the infinite geometric 

series 

a, ar, ar^^ ar^j ar^^ . • ., 

where r is numerically less than unity. The sum of the first n 
terms of this series is, by 6, p. 1, 

„ a(l — r") a af 

^' — Trr=i~r-rrr- 

If we now suppose n to increase without limit, the first fraction 

* Such a result is sometimes^ for the sake of brevity, called the gum of the series ; hut the 
student must not forget that 2 is iu>t the sum but the limit of the tumj as the number of terms 
increases without limit. 



220 DIFFERENTIAL CALCULUS 

on the rightrhand side remains unchanged, while the second 
approaches zero as a limit. Hence 

limit « ^ 
n = ao * 1 — r 

a perfectly definite number in any given case. 

In Case II the infinite series is said to be rumconvergent,* Series 
under this head may be divided into two classes. 

First class. Divergent series, in which the sum of n terms 
increases indefinitely in numerical value as n increases without 
limit; for example, the series from which we get 

/S.=l+2 + 84---- + w. 

As n increases without limit, S^ increases without limit and 
therefore the series is divergent 

Second class. Oscillating series, of which 

^,=1-1 + 1-1+. .. + (-1)-^ 

is an example. Here S^ is zero or unity according as n is even 
or odd, and although S^ does not become infinite as n increases 
without limit, it does not tend to a. limit, but oscillates. It is 
evident that if all the terms of a series have the same sign the 
series cannot oscillate. 

Since the sum of a converging series is a perfectly definite num- 
ber, while such a thing as the sum of a nonconvergent series does 
not exist, it follows at once that it is absolutely essential in any 
given problem involving infinite series to determine whether or 
not the series is convergent. This is often a problem of great 
difficulty, and we shall consider only the simplest cases. 

149. Existence of a limit. When a series is given we cannot in 
general, as in the case of a geometric series, actually find the num- 
ber which is the limit of S^. But although we may not know 
how to compute the numerical value of that limit, it is of prime 
importance to know that a limit does exist, for otherwise the series 
may be nonconvergent. When examining a series to determine 
whether or not it is convergent, the following theorems, which we 
state without proofs, are found to be of fundamental importance.! 

* Some writers nse divergent as equivalent to tionamvergent, 
t See Osgood's Introduction to Infinite Seriett pp. 4, 14, 04. 



SERIES 221 

Theorem I. If S^ is a variable that always increases as n increases^ 
hut always remains less than some definite fia^d number A^ then as 
n increases without limitf S^ will approach a definite limit which is 
not greater than A. 

Theorem II. If S^ is a variable that always decreases as n increases^ 
but always remains greater than some definite fixed number B, then 
as n increases without limits S^ will approach a definite limit which 
is not less than B, 

Theorem III. The necessary and sufficient condition that S^ shall 

approach some definite fixed number as a limit as n increases without 

limit is that ,. .. 

limit .^ _S)=0 

for all values of the integer p. 

150. Fundamental test for convergence. Summing up first n and 
then n -f y terms of a series, we have 

{A) ^« = Wi + w, -f Mj + . . . 4- u^. 

(B) /S,+^ = tti-fw,-f Ws + '-' + Wn + w.+i + '-' + ^.+p- 
Subtracting (-4) from (J9), 
{(^) ^n^p - 'S, = tt,^.i 4- u^^2 + • • • + w„+p. 

From Theorem III we know that the necessary and sufficient 
condition that the series shall be convergent is that 

for every value of p. But this is the same as the left-hand member 
of ((7); therefore from the right-hand member the condition may 
also be written 

Since (B) is true for every value of p, then letting je? = 1, a 
necessary condition for convergence is that 

or, what amounts to the same thing. 



222 DIFFERENTIAL CALCULUS 

Hence, if the general (or nth) term of a series does not approach 
zero as n approaches infinity, we know at once that the series is 
nonconvergent and we need proceed no further. However, (H) is 
not a sufficient condition, that is, even if the wth term does approach 
zero we cannot state positively that the series is convergent ; for, 
consider the harmonic series 

.111 1 

' 2 3 4 'n 



Here j''"'* («,) = »■»" (-\ = 0, 



that is, condition (E) is fulfilled. Yet we may show that the har- 
monic series is not convergent by the following comparison: 

{F) l+J + [J + J] + [J+J + ^+|] + [i + ... + ^] + .... 
(«) i + i+[t + i] + [i + i + i + i] + [^ + - + ^]+--. 

We notice that every term of {O) is equal to or less than the 
corresponding term of (-F), so that the sum of any number of the 
first terms of {F) will be greater than the sum of the corresponding 
terms of (6r). But since the sum of the terms grouped in each 
bracket ui.{0) equals ^, the sum of (6r) may be made as large 
as we please by taking terms enough. The sum ((?) increases 
indefinitely as the number of terms increases without limit ; hence 
((7), and therefore also (^), is divergent. 

We shall now proceed to deduce special tests which as a rule 
are easier to apply than the above theorems. 

151. Comparison test for convergence. In many cases, an example 
of which was given in the last section, it is easy to determine 
whether or not a given series is convergent by comparing it term 
by term with another series whose character is known. Let 

{A) Uj4-Wj4-w,4. ... 

he a aeries of positive terms which it is desired to test for convergence. 
If a series of positive terms already knotun to be convergent^ namely^ 

can be found whose terms are never less than the corresponding terms 



SERIES 228 

in the series (A) to be tested^ then {A) is a convergent series and its 
sum does not exceed that of (B). 

Proof, Let «« = t^i 4- Wj 4- w, h 1- w^, 

and ^n = «i + «a + «« 4- • • -f «, ; 

and suppose that ,. . 

« — «« '^M = -^• 

fl = oo " 

Then, since 'Sn<^ ^^^ *ii ~ '^h'» 

it follows that s^ < A. Hence, by Theorem I, p. 221, s^ approaches 
a limit ; therefore the series (-4.) is convergent and the limit of its 
sum is not greater than A. 

Ex. 1. Test the series 

(C) i+i + l + l + 14..... 

^ ' 22 3» 4* 6* 

Solution. Each term after the first is less than the corresponding term of the 

geometric series 

1111 
(D) 1 + - + -+- + - + •••, 

which is known to be convergent (p. 218) ; hence (C7) is also convergent. 

Following a line of reasoning similar to that applied to (A) and 
(-B), it is evident that, if 

(JE) w^ ^. w^ 4. Wj 4. . . . 

is a series of positive terms to be tested which are never less than the 
corresponding terms of the series of positive terms, namely , 

(F) 5^4.6,4.63 + ..., 

known to be divergent, then (U) is a divergent series. 

Ex. 2. Test the series ^ . ^ 

V^ V3 V4 

Solution, This series is divergent since its terms are greater than the corre- 
sponding terms of the harmonic series 

, 1 1 1 

1 + - + - + » 

.284 ' 

which is known (p. 222) to be divergent. 

Ex. 3. Test the following series for different values of p. 

(G) i4JL + i- + i-4.. . 

2p Zp ^P 



h* 



a 
1 



224 DIFFERENTIAL CALCULUS 

SobUion, Grouping the terms, we have, when p > 1, 

2p 8i» 2p 2p^2p 2''-i' 

4p S** 61' 7p 4p 4p 4^ 4p 4p \2p-i/ 

UP Sp Sp Sp SP bP Sp Sp Sp 8p \2i»-i/ 
and BO on. Construct the series 

When p > 1, series (H) is a geometric series with the common ratio less than 
unity, aitd is therefore convergent. But the sum of {G) is less than the sum of 
{H), as shown by the above inequalities ; therefore (€f) is also convergent. 

When p = 1, series {G) becomes the harmonic series which we saw was 
divergent, and neither of the above tests applies. 

When p < 1, the terms of series {G) vnll, after the first, be greater than the 
corresponding terms of the harmonic seiies ; hence (G) is divergent. 

152. Cauchy's ratio test for convergence. Let 

(A) u^ + u^^u^^.>. 

be a series of positive terms to be tested. 

Divide any general term by the one that immediately precedes 

it, i.e. form the test ratio -^^ 

As n increases without limit, let ^^ -2±i = p. 

W= QO w ' 

I. When p<l. By the definition of a limit (§ 29, p. 19) we 



can choose n so large, say w = w, that when n^m the ratio 



u 



w+i 



u 



shall differ from p by as little as we please, and therefore be less 
than a proper fraction r. Hence 

and so on. Therefore, after the term w„, each term of the series 
(A) is less than the corresponding term of the geometrical series 

But since r < 1, the series (5), and therefore also the series (-4), 
is convergent.* 

* When examining a series for convergence we are at liberty to disregard any finite number 
of terms ; the rejection of such terms would aif ect the value but not the exUtence of the limit. 



SERIES 225 

II. When p>l {or p= c»). Following tho same line of reason- 
ing as in I, the series (A) may be shown to be divergent. 

III. When p = 1, the series may be either convergent or diver- 
gent; that is, there is no test. For, consider the series ((7) on 
p. 223, namely, 



2" 8' 4f tif {n + ly 

The test ratio is ^^> = (-^\ = (l ^-Y; and 

limit /m.+A_ limit A _1 V ,.., ^, . 

Hence p = 1 no matter what value p may have. But on p. 224 
we showed that 

when jt? > 1, the sei ies converges, and 
when JE? < 1, the series diverges. 

Thus it appears that p can equal unity both for convergent and 
for divergent series, and the ratio test for convergence fails. There 
are other tests to apply in cases like this, but the scope of our book 
does not admit of their consideration. 

Our results may then be stated in compact form as follows: 

Ghiven the series of positive terms 

t^i 4- Wg 4- Wj -f . . . -f w„ -f w^+i + • • •; 
find the limit ^^^^ (^^^ = P- 

I. When p < 1,* the series is convergent. 
II. When p > 1, the series is divergent. 
III. When /> = 1, there is no test. 

Ex. 1. Test the following series for convergence : 

c = 1 + i + ,4 + ,4 + ri + • • • + -^ + .- + •••. 
1 [2 [3 [4 \n-\ \n 



* It is not enongh that um+\/un becomes and remains less tlian unity for all yalues of n, but 
this test requires tbat tbe limit of u»+i/u«i sball be less tban unity. For instance, in the case of 
the harmonic series this ratio is always less than unity and yet the series diverges as we haye 
seen. The limit, however, is not less than unity but equals unity. 



226 DIFFERENTIAL CALCULUS 



8olvti4m. The nth term is ; therefore 

n — 1 



Umit (^^!L±l\= limit / ^ \ ^ limit AlL:zl \ _ limit /l\-or=p) 
n = oo\ t4„ / n = «l 1 I n = »\ [n / n = co\n/ ^ '' 

\[nEi/ 

and by I, p. 225, the series is convergent. 

\l_ \2_ \S_ 

Ex.2. Test the series t:: + r^ -^ tzi + " ' - 

10 102 io» 

\J1 
Solution. The nth term is here — ; therefore 

10" 

limit /l??i±i\ ^ limit AlL±l ^ 10»\ _ n^a / n + l \ _. ^ 
n = oo\ Un / n = ao\iOi»+i [n / n = oo\ IQ / ' 

and by n, p. 226, the series is divergent. 

Ex. 3. Test the series 

(O — + — + — +•••. 

^ ' 123. 46. 6 

Solution, Here the nth term is ; therefore 

(2n-l)2n 

limit /«»-n\^ limit [ (2 n ~ 1) 2 n -j ^^ 
n = oD\ M, / n = <» L(2n + l)(2n4-2)J 

This gives no test (III, p. 225). But if we comi>are series (C) with (G), p. 223, 
making p = 2, namely, 

(D) 1 + — + — H h • • •, 

we see that (C) must be convergent since its terms are less than the corresponding 
terms of (2>), which was proven convergent. 

153. Alternating series. This is the name given to a series 
whose terms are alternately positive and negative. Such series 
occur frequently in practice and are of considerable importance. 

Jf Wj — w, 4- 1^, — w^ H 

is an alternating series whose terms never increase in numerical value^ 

and if ^"^'l u^ = 0, 

then the series is convergent. 

Proof. The sum of 2 n (an even number) terms may be written 
in the two forms 

(^) 'S,, = (i^i-t^,) + (t4,-w,) + K-We)4-----fK.-i-0»or, 



SERIES 227 

Since each difference is positive (if it is not zero, and the assump- 
tion ^*^*^ ^« = excludes equality of the terms of the series), 
series (A) shows that S^^ is positive and increases with w, while 
series (B) shows that S^^ is always less than u^; therefore by Theo- 
rem I, p. 221, 5,, must approach a limit less than u^ when n 
increases, and the series is convergent. 

Ex. 1. Test the alternating series 1 1 1 . 

^ 2 8 4 

Solution, Since each term is leas In numerical value than the preceding one, and 

limit (^)^ limit /I \o. 
n = oo^ ' n = oo\n/ 
the series is convergent. 

154. Absolute convergence. A series is said to be absolutely* or 
unconditionally convergent when the series formed from it by mak- 
ing all its terms positive is convergent. Other convergent series 
are said to be not absolutely convergent or conditionally convergent. 
To this latter class belong some convergent alternating series. For 
example, the series 

2» 3« 4* 6* 
is absolutely convergent since the series ((7), p. 228, namely, 

^2* 3« 4* 6* 
is convergent. The series 

2^3 4^6 
is conditionally convergent since the harmonic series 

^1111. 

^2^3^4^6^ 
is divergent (p. 222). 

A series with terms of different signs is convergent if the series 

deduced from it by making all the signs positive is convergent. 

The proofs of this and the following theorem are omitted. 

*Tbe tensB of the new series are the Dumerical (absolute) yaluee of the tenns of the glyeo 
series. 



228 



DIFFERENTIAL CALCULUS 



Without placing any restriction on the signs of the terms of 
the series, the tests given on p. 225 may be stated in the following 
more general form : 

Ghiven the series 

Wi + W« + W, 4- «^4 + • • • -f Wn + W« + 1 + • • • ; 

calculate the limit ( -=^^ ) = />, 



limit /!!5iL±i\ = 



L When\p 
II. When\p 



< 1, the series is absolutely convergent. 
> 1, the series is divergent. 
III. When |p| = 1, there is no test. 

155. Power series. A series of ascending integral powers of a 
variable, say x^ of the form 

{A) a^ -f a^x -f- a^ -f a^ H — , 

where the coefficients a^, a^, a,, . • • are independent of :r, is called 
a power series in rr. Such series are of prime importance in the 
further study of the Calculus. 

In special cases a power series in x may converge for all values 
of rr, but in general it will converge for some values of x and be 
divergent for other values of x. We shall examine {A) only for 
the case when the coefficients are such that 



limit / a»+A 
n^co\ a. ) 



A 



where Z is a definite number. In {A) 
lunit '- ^ ^~=^ '- -"-^^ 



n 



unit /^,^i\ limit /5i±l^\= li™i^ /^n^A ^r^ia: 



Referring to tests I, II, III, we have in this case 

p = Lx, 
and hence the series (A) is 

I. Absolutely convergent when \Lz\<ly or \x\< 



lies between 



1 
L 



and -f 



1 
L 



1_ 
L 



i.e. when x 



II. Divergent when \Lx\>\^ or \x\> 



than — 



L 



or greater than + 



1 
L 



L 



i.e. when x is less 



SERIES 



229 



1 
L 



; i.e. when z = dt 



1 
L 



III. No test when \Lx\=:ly or \x\ = 

Note. When Z = 0, it is evident from I (p. 228) that the power 
series is absolutely convergent for all finite values of 3c 

Ex. 1. Test the series 

SolutUm. The series formed by the coefficients is 



X h 

2« 3« 4« 



22 ^ 32 42 ^ 



Here 



limit /«»±i\ ^ limit r '^^ 1:= limit r/j l_Vl=-l/' = i;\ 



L\ \-l 



= 1. 



By I the series is absolutely convergent when x lies between — 1 and + 1* 

By II the series is divergent when x \a less than — 1 or greater than + 1. 

By III there is no test- when x = ±l. But in either case (B) is convergent from 
the first theorem under § 164, p. 227, since (D), p. 226, was proved convergent. 

The series in the above example is said to have [— 1, 1] as the irUervcU of conver- 
gence. This may be v^ritten — 1 < x < 1, or indicated graphically as follows : 



-1 



Show that the following nine series are convergent. 



1. l + i + i^- •••. 
12 ^ 22 ^ 32 ^ 

2. 1 + 1 + 1 + 1+.. 
22228 2*^ 

4 1 + Ll? + LA:^ + 

' 3 36 3.6.9 



o. — I — + — + •••• 

\^ [k \i 

2V2 8v3 4V4 
7 1-^ + ^ ^ + ^ 

8. 1-1. 1 + 1. 1-1. 1 + . 
2 2233 2* 42* 



9. 



' + ' 



log 2 log 3 log 4 
Show that the following three series are divergent. 



10.1 + 1 + 1+.... 

12 1 + 1 + 2.1 + 3. 1+4 



ll.ii + i?+-l4 + 

10 102 108 



230 DIFFERENTIAL CALCULUS 

For what values of the variable are the following Graphical representations of 

series convergent ? *^**™^ ^' oonvergenoe .• 



[2 • [3 



1» 3a 62 

Ans, AU values of a. 



® 
-1 41 



13. l + x + x* + «•••. Ana, -l<x<l. Q) I ® 

-1, +1 



14. X -- + --- + .... ^TW. -l<x<l. 
2 3 4 



16. X + X* + aC» + x^' + • • • . -4ns. — 1< x < 1. m I o 

-¥ +1 

16. x + -^4--^ + - •. Am. --1<x<1. ' ■ i Q 

V2 V3 ^ 5 +1 



X^ x' —00 I ^<o 

17. 1 + X + — + — 4----. An8. All values of as. ^ 







^ ^ ^ — eo I -f-«o 

18. 1 1 !-•••. An8. AU values of $. < I > 

12 li li 5 

^« ^ «aT -co I -f«o 

19. 0-i- + i--f- + .... ^na. All values of 0. ^ I » 

15 15 II *» 

^^ sin o sin 3 o sin 5 a -«> | +« 
^O. — ■ — — — — -p 



i 



rt- C08X C08 2X . C08 3x . , _ I -^ 

21. \- — h — 1 Ans. X > 0. d i > 

Bint. Neither the sine nor ooeine can exceed 1 numerically. 

22. l4.xloga-f5^ + ?^ + .... Z^ I i? 

[2 [3 

Ana. All values of x. 

Ill 1 +00 

23. + ;; - + ; 1 + --. Ans. X > 1. 1 ® » 

1 + X 1 + X2 1 + X« 0+1 



^. 1 x8 1.3 x6 , 1.36 xi . ^ 

24. X H H h h • • • . (D I ® 

2 3 2-462.4.67 -1 +i 

Ana. — 1 < X < 1. 

• End points that are not included in the interval of oonvergenoe have drcles drawn about 
them. 



CHAPTER XX 

EXPANSION OF FUNCTIONS 

156. Introduction. The student is already familiar with some 
methods of expanding certain functions into series. Thus, by the 
Binomial Theorem, 

(A) (a -f a:)* = a* + 4 a"a: 4- 6 aV -f 4 oa?" -f a^, 

giving a finite power series from which the exact value of (a -f- x)* 
for any value of x may be calculated. Also by actual division, 

1 — X \l — xy 

we get an equivalent series, all of whose coefficients except that 
of of are constants, n being a positive integer. 

Suppose we wish to calculate the value of this function when 
X = .5, not by substituting directly in 

1 
but by substituting x = .5 in the equivalent series 

Assuming n = 8, ((7) gives for a; = .5 

(D) . zA— =1.9921875 + .0078125. 

1 — a: 

If we then assume the value of the function to be the sum of 
the first eight terms of series ((7), the error we make is .0078126. 
However, in case we need the value of the function correct to two 
decimal places only, the number 1.99 is as close an approximation 
to the true value as we care for since the error is less than .01. 
It is evident that if a greater degree of accuracy is desired, all we 
need to do is to use more terms of the powe^ series 

{H) l+a;-fa:*4-a:*H 

231 



232 • DIFFERENTIAL CALCULUS 

Since, however, we see at once that 

2, 



Li-J.-." 



there is no necessity for the above discussion, except for purposes 
of illustration. As a matter of fact the process of computing the 
value of a function from an equivalent series into which it has 
been expanded is of the greatest practical importance, the values 
of the elementary transcendental functions such as the sine, cosine, 
logarithm, etc., being computed most simply in this way. 

So far we have learned how to expand only a few special forms 
into series; we shall now consider a method of expansion appli- 
cable to an extensive and important class of functions and called 
Taylor's Theorem. 

157. Taylor^s Theorem* and Taylor^s Series. Replacing J by a: 
in (j&), p. 169, the extended theorem of the mean takes on the form 

(59) A^) = Aa) + ^^^^r(a) + i^^/Fr(a) + ^^^^^fnf^a) + • • • 

Li I* Is 

where x^ lies between a and a*. (59), which is one of the most far- 
reaching theorems in the Calculus, is called Taylor^B Theorem. 
We see that it expresses /(a:) as the sum of a finite series in {x — a). 

(x — aY 
The last term in (59), namely, ^— ; — -f^^Mt is sometimes called 

|n 

the remainder in Taylor's Theorem after n terms. If this remainder 
converges towards zero as the number of terms increases without 
limit, then the right-hand side of (59) becomes an infinite power 
series called Taylor^ Series,^ In that case we may write (59) in 
the form 

(60) f(x) = f(a) + ^^^^r{a) + <^^/rf (a) + ^^^ f^a) + • • ., 

LL L* 15 

and we say that the function has been expanded into a Taylor^ s Series. 
For all values of x for which the remainder approaches zero as n 
increases without limit, this series converges and its sum gives the 

* AIbo known as Taylor*i Formula. 

t Published by Dr. Brook Taylor (1685-1731) in his Methodut IncremetUorum^ London, 1715. 



EXPANSION OF FUNCTIONS 233 

exact value of /(x), because the difference (= the remainder) between 
the function and the sum of n terms of the series approaches the 
limit zero (§ 30, p. 21). 

On the other hand, if the series converges for values of x for 
which the remainder does not approach zero as n increases without 
limit, then the limit of the sum of the series is not equal to the 
function /(a:). 

The infinite series (60) represents the function for those values of x 
and those only for which the remainder approaches zero as the num- 
ber of terms increases without limit. 

It is usually easier to determine the interval of convergence 
of the series than that for which the remainder approaches zero ; 
hut in simple cases tlie two intervals are identical (see footnote, 
p. 236). 

When the values of a function and its successive derivatives are 
known for some value of the variable, as 2; = a, then (60) is used 
for finding the value of the function for values of x near a, and (60) 
is also called the expansion of f{x) in the vicinity of x=za. 

Ex. 1. Expand log x in powers of (x — 1). 
Solution. f(x) = log X, /(I) = ; 

r(x) = -, /'(1) = 1; 

X 
X' 



Substituting in (60), logx = x -1 -i(x - l)« + i(x - 1)» . Ana. 

Ttiis converges for values of x between and 2 (§ 166, p. 228) and is the expan- 
sion of log X in the vicinity o/x = 1, the remainder converging to zero. 

When a function of the sum of two numbers a and x is given, 
say f{a -f x), it is frequently desirable to expand the function into 
a power series in one of them, say x. For this purpose we use 
another form of Taylor's Series, gotten by replacing a; by a -f a: in 
(60), namely, 

(61) f(a + x)= f(a) + jf /'(a) + ,^/"(a) + %rHa) + • • •. 

11 15 Iff 



234 DIFFERENTIAL CALCULUS 

Ex. 1. Expand sin (a + x) in powers of x. 

SoiutioTL Here f{a + x) = Bin (a + x) ; 

hence, placing x = 0, 

/(a) = sin a, 
/'(a) = COB a, 
/"(a) = - sin a, 
/'"(a) = - COS a, 

Substituting in (61), 

X x^ x^ 

sin (a 4- x) = sin a + - COB a — -- sin a — -- COS a + • • • . J.1W. 



EXAMPLES* 

1. Expand e* in powers of X- 2. Ans. c* = c» + ^(x - 2) + — (x - 2)«+ •• •. 

I? 

2. Expand x* — 2x^ + 5x — 7 in powers of x — 1. 

Ana. -3 + 4(x-l) + (x-l)«+(x-l)". 

3. Expand 3v*~14y + 7in powers of y — 3. 

Am. -8 + 4(y-3) + 3(y-3)«. 

4. Expand 5z3 + 7z + 3in powers of z — 2. 

Ana. 37+27(«-2) + 6(«-2)«. 

6. Expand cos (a + x) in powers of x. as 

Ana. cos(a + x) = cos a — xsin a — — cos a H — sin a + • • •. 

|2 [3 

6. Expand log (x + h) in powers of x. 2 ^a 

Ana. log(x + A) = log;i + ^-^ + — + .... 

7. Expand tan (x + h) in powers of h. 

Ana. tan (x + A) = tan x + A sec^ x + A'sec* x tan x + • • • . 

8. Expand the following in powers of h. 

(a) (x + A)'> = x" + nx''-'^ + ''^^"'^^ x-«ft8 + ^^'*"'|^^'^^^^ x"-8A8 + "-> 

[2 [3 



(l,)e«+* = c«(l + A + | + ^ + ...). 



[2 li 

158. Maclaurin's Theorem and Maclaurin's Series. A particular 
case of Taylor's Theorem is found by placing a = in (59), p. 232, 
giving 

(62) fix) = /(O) + ^ /'(O) + ^ fff(0) + ^ /'"(O) + . . . 

Lt I* lil 

+ if— r /<"-"' (O) + ^ /^"> («!). 

* In these examples we assume that the functions can be dereloped Into a power series. 



EXPANSION OF FUNCTIONS 236 

where x^ lies between and x. (69) is called Maclaurin's Theorem. 
The right-hand member is evidently a series in a; in the same sense 
that (59)9 p. 232, is a series in x — a. 

Placing a = in (60), p. 232, we get MaclaurirCs Series^* 

(63) f(x) =/(0) + j^/'(0) + 2!/fr(0) + ^/'"(O) + •.., 

Li I* 12 

a special case of Taylor's Series that is very useful. The state- 
ments made concerning the remainder and the convergence of 
Taylor's Series apply with equal force to Maclaurin's Series, the 
latter being merely a special case of the former. 

The student should not fail to note the importance of such an 
expansion as (63). In all practical computations results correct 
to a certain number of decimal places are sought, and since the 
process in question replaces a function perhaps difficult to calcu- 
late by an ordinary polynomial with constant coefficients^ it is veiy 
useful in simplifying such computations. Of course we must use 
terms enough to give the desired degree of accuracy. 

In the case of an alternating series (§ 153, p. 226) the error 
made by stopping at any term is numerically less than that term, 
since the sum of the series after that term is numerically less than 
that term. 

Ex. 1. Expand cos x into an infinite power series and determine for what values 
of X it converges. 

Solution, Differentiating first and then placing x = 0, we get 

f(x) = cos X, /(O) = 1, 

/'(x) = ~sinx, /'(0) = 0, 

r(x)=-cosx, r(0)=-l, 

/"' (X) = sin X, /'" (0) = 0, 

/^▼(x) = cosx, /iv{0) = l, 
/^(x) = -sinx, ' /^(0) = 0, 

/Ti(x) = - cosx, /^(O) = - 1, 
etc., etc. 

Suhstitnting in (63), 

X* X* a^ 

(A) C08X = 1--+--- + .... 

• Named after Colin Maclanrln (1698-1746), being first published in his Treatise o/Fluxiant, 
Edinbuigh, 1743. The Series is really due to Stirling (1091M770). 



236 DIFFERENTIAL CALCULUS 

Comparing with Ex. 18, p. 230, we see that the series converges for ail values 
of as. 

In the same way for sin x. 

gj8 2^ 2"7 

(B) «nx = «-- + --- + .... 

which converges for all values of x (Ex. 10, p. 230).* 

Ex. 2. Using the series {B) found in the last example, calculate sin 1 correct to 
four decimal places. 

SoltUion, Here x = 1 ; that is, the angle is expressed in circular measure (see 
second footnote, p. 17). Therefore, substituting x = 1 in (B) of the last example, 

Summing up the positive and negative terms separately, 

1 = 1.00000- . • .- = 0.16667. . • 

i = 0.00833- • . .- = 0.00019- . • 

ti L7 

1.00833' - 0.16686--' 

Hence sin 1 = 1.00833 - 0.16686 = 0.84147 - - • , 

which is correct to four decimal places since the error made must be less than 
— } i.e. less than .000003. Obviously the value of sinl may be calculated to any 

desired degree of accuracy by simply including a sufficient number of additional 
terms. 

EXAMPLES 

Verify the following expansions of functions into power series by Maclaurin^s 
Series and determine for what values of the variable they are convergent. 

jpfl jg' x^ 

1. ^ = 1 + aj + — -f — - + — i . Convergent for all values of x. 

L± l£ Lz 

x^ x^ sfi x' 

2. co8x = 1 —,-- + ■- — — + — •. Convergent for all values of X. 

[2 [4 |6 [8 

• Since here /('•)(x)=ain (x + ^) and /(»)(xt)«»ain (x^ + — ) , we have, by subetitating in the 
lai»t term of (62), p. 234, ^^ 2 / \ 2/ 

remainder = — Bin I XiH )• 0<Xi<ar 

[n \ 2 / . 

But sin (xi + — ) can never exceed unity, and from Ex. 17, p. 230, Wmit — =o for all values 
\ 2 / w=qo|n 

of X. Hence ijmif ar" . / nn\ ^ 

limit _8in(x, + ^)=0 
n=ao[n \ * 2/ 

for all ralues of x ; that is, in this case the limit of the remainder i» for all valuea of x for 
which the series converges. This is also the case for all the functions considered in this book. 



EXPANSION OF FUNCTIONS 



287 



3. «. = i + xlog, + ^l^ + ^i^ + 



12 



[8 



4. sin ke = KX h !-■••. 

13 16 II 

^ 1 /^ V X' X» 3C* X* 

6. log (1 + x) = X h • • 

'^^^ 23 46 

« , .- . X* x« X* X* 

7. log (1 — x) = — X 

8. arc sin X = X H h h • • • . 

23 245 

x' x^ x^ sfi 

9. arc tan x = x 1 — -\ •.-. 

3 6 7 9 

10. 8in2x = x«-- — + — +•••. 

L3 [6 

11. erfii«fr = i^ 0^r__zL-|- .... 

2 8 

9" 4^ 8^ 

12. c«8in^=^+^+^- ---^ . 

3 ^ 



Conyergent for all values of x. 

Convergent for all values of x, 
k being any constant. 

• 

Convergent for all values of x, 
k being any constant. 

Convergent if — 1 < x < 1. 
Convergent if ~ 1 < x < 1. 
Convergent if — 1 < x < 1. 
Convergent if — 1 < x < 1. 
Convergent for all values of x. 
Convergent for all values of 0. 
Convergent for all values of 0. 



13. Show that log x cannot be expanded by Maclaurin^s Theorem. 

Compute the values of the following functions by substituting directly in the 
equivalent power series, taking terms enough until the results agree with those 
given below. 

14. 6 = 2.7182... 

SohUUm. Let x = 1 in series of Ex. 1 ; then 

c = l + H--H 1 !-—+•••. 



First term 


= 1.00000 




Second term 


= 1.00000 




Third term 


= 0.50000 




Fourth term 


= 0.16667.. 




Fifth term 


= 0.04167 . . 




Sixth term 


= 0.00833.. 




Seventh term 


= 0.00139.. 




Eighth term _ 


= 0.00019 . . 


•, etc. 


Adding, e 


= 2.71826.. 


Ans, 



[DiTiding third term by 3.] 
[Dividing fourth term by 4.] 
[IMriding fifth term by 5.] 
[Dividing sixth term by 6.] 
[Dividing seventh term by 7.] 



15. arc tan (I) = 0.1973 • > . ; use series in Ex. 9. 

16. cos 1 = 0.5403 • • . ; use series in Ex. 2. 



17. cos 10° = 0.9848 • • • ; use series in Ex. 2. 



288 DIFFERENTIAL CALCULUS 

18. sin 7 = 0.7071 • • • ; iwe series (B), p. 28a 

4 

19. sin .6 = 0.4794 • • • ; use series (B), p. 236. 

20. c9 = 1 4- 2 + - + .^ + • • • = 7.3891. 

[2 [3 

169. Computation by series. 

I. 2%e computation of ir hy series. 
From Ex. 8, p. 287, we have 

arc sin x^=X'\ h 1- . . •. 

^2.8 2.4.5 2.4.6.7 

Since this series converges ♦ for values of x between — 1 and + 1, 
we may let a; = i, giving 

6 2^2 3V2y ^2.4 5V2y ^ • 

or, 7r = 3.1415 ... 

Evidently we might have used the series of Ex. 9, p. 237, instead. 
Both of these series converge rather slowly, but there are other 
series, found by more elaborate methods, by means of which the 
correct value of tt to a large number of decimal places may be 
easily calculated. 

II. The computation of logarithms by series. 

Series play a very important r81e in making the necessary calcu- 
lations for the construction of logarithmic tables. 
From Ex. 6, p. 287, we have 

(A) log(l + x) = z-- + --- + ^-.... 

This series converges for z = 1, and we can find log 2 by placing 
X = 1 in (A), giving 

log2 = l-^ + J-} + J-J+.... 

But this series is not well adapted to numerical computation, 
because it converges so slowly that it would be necessary to take 

• We asaume that it oonverges to the correct value. 



EXPANSION OF FUNCTIONS 239 

1000 tenns in order to get the value of log 2 correct to three deci- 
mal places. A rapidly converging series for computing logarithms 
will now be deduced. 

By the theory of logarithms, 

{B) logl±^ = log(l + n:)-log(l~n:). 8, p. 2 

1 — X 

Substituting in (B) the equivalent series for log (1 -fa;) and 
log(l — a;) found in Exs. 6 and 7 on p. 237, we get* 

which is convergent when x is numerically less than unity. Let 

(2>) --il^ = --:, whence x^ "" , 

1 — x N M-\-N 

and we see that x will always be numerically less than unity for 
all positive values of M and N. Substituting from (2>) into ((7), 
we get 

(E) log^ = logJf-logJ\r 

a series which is convergent for all positive values of M and N ; 
and it is always possible to choose M and ^ so as to make it con- 
verge rapidly. 

Placing Jf = 2 and JV = 1 in {U)^ we get 

l°g2 = 2[l-Hl.i + l.i, + l.i + ...]=0.69314718.... 

[since log N^ log 1-0, and ~ « - • 1 

L Jo + iV S J 

Placing Jf = 8 and iV= 2 in (J?), we get 
logS =log2 + 2 n + i.i + ^.i + .. ."I =1.09861229.... 

*The student shonid notice that we have treated the series as if they were ordinary 
stuns, but they are not ; they are limits of sums. To justify this step is beyond the scope of 
this book. 



240 DIFFERENTIAL CALCULUS 

It is only necessary to compute the logarithms of prime numbers 
in this way, the logarithms of composite numbers being then found 
by using theorems 7-10, p. 2. Thus, 

log 8 = log 2» == 3 log 2 = 2.07944154 • • ., 
log 6 = log 8 + log 2 = 1.79175947 .... 

All the above are Naperian or natural logarithms^ i.e. the base is 
e = 2.7182818. If we wish to find Brigg%^ or common logarithmsy 
where the base 10 is employed, all we need to do is to change the 
base by means of the formula 

Thus, w 9=.igg^=0'693...^ 

^'' log. 10 2.802... 

In the actual computation of a table of logarithms only a few of 
the tabulated values are calculated from series, all the rest being 
found by employing theorems in the theory of logarithms and 
various ingenious devices designed for the purpose of saving 
work, 

EXAMPLES 

Calculate by the methods of this article the following logarithms. 

1. log^S =1.6094.... 3. loge 24 = 3.1781.... 

2. logelO = 2.3026 ... 4. logio 5 = 0.6990 .-• . 

160. Approximate formulas derived from series. In the two pre- 
ceding sections we evaluated a function from its equivalent power 
series by substituting the given value of 2; in a certain number of 
the first terms of that series, the number of terms taken depend- 
ing on the degree of accuracy required. It is of great practical 
importance to note that this really means that we are considering 
the function as approximately equal to an ordinary polynomial with 
constant coefficients. For example, consider the series 

{A) Binx = x-| + |-^ + .... 

This is an alternating series for both positive and negative values 
of X. Hence the error made if we assume sin a; to be approximately 



EXPANSION OF FUNCTIONS 241 

equal to the sum of the first n terms is numerically less than the 
(n -f l)th term, § 153, p. 227. For example, assume 

(B) sin x = x^ 

and let us find for what values of x this is correct to three places 
of decimals. To do this, set 



(0) 






< .001. 



This gives x numerically less than V.006 (= .1817); that is, {B) 
is correct to three decimal places when x lies between -{- 10° .4 and 
- 10°.4. 

The error made in neglecting all terms in (A) after the one in 
i"~' is given by the remainder, (62), p. 234, 

m E = ^&\x,); 

\n 

hence we can find for what values of 2: a polynomial represents the 
function to any desired degree of accuracy by writing the inequality 

(E) \E\ < limit of error, 

and solving for a-, provided we know the maximum value off^*\x^). 
Thus, if we wish to find for what values of x the formula 

(F) sin a; = a: — — 

i; correct to two decimal places (i.e. error <. 01), knowing that 
|/«^>(a:i)|<l, we have from (2>) and (H), 

<M; that is,la:|<'V^; or, |a:|<l. 



120 



a:" 
Therefore x-r-^ gives the correct value of sin x to two decimal 

D 

places if |a^|< 1? i.e. if x lies between + 57° and — 57°. This agrees 
with the discussion of (A) as an alternating series. 

Since in a great many practical problems accuracy to two or three 
decimal places only is required, the usefulness of such approximate 
formulas as {B) and (F) is apparent. 



242 DIFFERENTIAL CALCULUS 

Muygens^ approximation to the length of a circular arc. 

Let S denote the length of the required arc, C its chord, c the 

chord of half the arc, and B the radius of the circle. The circular 

measure of the whole angle being 

S 

£ 1 

, . S 2 . S 2 

we have sin g-g = ;^' ^^^Ib^H' ^^' 

C . S c . S 
= sin--— 1 •r--=sin 




2Ii 2R 2B ^B 

Developing the right-hand members of the last two equations 
into power series in S by (68), p. 235, we get 

C 8 S" . S^ . 



(fl^) 



2B 2B 1 28. 2»^^ 12. 5. 2*^ 



\ f OJ? AT? 10Q0«J?8'' 



2R ^B l-2.3.2«^» • 1.2...6.2"^ 

Multipl3dng {H) by 8 and then subtracting ((?) from {H) in order 
to eliminate the term containing S\ we have approximately 

8g-(7 ^3^ 3 S^ 

2B ~'2B 4 12 .62*^' 

8c-C „ S' 
or, — - — = 5 — 



3 7680 -K* 

Hence, for an arc equal in length to the radius, the error in 
taking 

is less than j^-q of the whole arc. For an arc of half the length of 
the radius the proportionate error is one sixteenth less, and so on.* 
In practice Huygens' approximation is generally used in the 
form 

{J) S = 2c + ^{2c-C). 

• For an angle of 30* the error is less than 1 In 100,000, for 46* leas than 1 in 20,000, and for 00 
less than 1 In OOOO 



EXPANSION OF FUNCTIONS 243 

This simple method of finding approximately the length of an 
arc of a circle is much employed. To find the approximate length 
of a portion of any continuous curve, divide it into an even number 
of suitable arcs, regarding the arcs as approximately circular. 

Ex. 1. Find by Huygens* approximation the length of an arc of 30^ in a circle 
whose radius is 100,000 ft. 

Solution. Here c = 2 B sin 7° 30', C = 2 B sin 16® ; but from tables of natural 
sines we get sin 7° 30^ = .1306268, sin 16° = .2688190. 

Substituting in {J), a =^52369.71. The true value, assumuig r = 3.1416926, is 
62369.88 ; hence the error is but .17 ft., or about 2 inches. 



EXAMPLES 

1. Draw the graphs of the functions x, x , z — — H — respectively, and 

compare them with the graph of sin x. L L l_ 

2. If d is the distance between the middle points of the chord c and the circular 
arc a, show that the error in taking 

_8^ 

. , ,, 82 d* ^ " * 3 s 

IS less than . 

3 a» 

161. Taylor's Theorem for functions of two or more variables. 
The scope of this book will allow only an elementary treatment 
of the expansion of functions involving more than one variable 
by Taylor's Theorem. The expressions for the remainder are 
complicated and will not be written down. 

Having given the function 

it is required to expand the function 

{B) /{x+h.y + k) 

in powers of h and k. 
Consider the function 

(C) f(x + U,y + kt). 

Evidently {B) is the value of {€) when < = 1. Considering (C) 
as a function of t, we may write 

(D) fix + ht,y + kt)= F(t), 



244 



DIFFERENTIAL CALCULUS 



which may then be expanded in powers of t by Maclaurin's Theo- 
rem, (62), p. 234, giving 

(H) F{t) = F{0) + tF^ (0) + ^ F''{0) + 1 F^''(0) + • • • . 

Let us now express the successive derivatives of F{t) with respect 
to ^ in terms of the partial derivatives of F{t) with respect to x and 
y. Let 

(F) a = x-\-ht, /3 = f/-\-kt; 

then by (49), p. 199, 
But from (F), 



j,n^.dFdadFdfi 
^ ^ da dt dfi dt^ 



-rr = h and -^szk; 
dt dt 



and since F{t) is a function of x and y through a and /8, 

dF^dFda , dF_dFdfi^ 
dx dadx dy~ dfi dy' 

or, since from (-F), — = 1 and — = 1, 

^ ' dx dy 



{I) 



dF dF J dF dF 
— = — and — = — • 
dx da dy d/3 



Substituting in (Cr) from (/) and (S), 

(J) 



^' dx dy 



Replacing F{t) by F'(f) in (J"), we get 



dx 



dy 



^^^kSl^^k 



dx 



dxdy 



[ dxdy^ dfj 



(K) 



... F"{t) = h'll+2hk?^ + ¥^^ 



dx' ■ dxdy 

In the same way the third derivative is 



dy". 



^"'(.) = ..g^3.«*^-^ + 3M'^ + *«0, 



and so on for higher derivatives. 



EXPANSION OF FUNCTIONS 246 

When t = 0, we have from (2>), (6?), (J'), (JT), (X), 

^(0)=/(^» y)» i-e« J^(0 ^ replaced \yjf\x, y), 

^ ' dx dy 
^•"(0) = A* ^ + 2 AA -^ + A* ^,, 

i?""(0) = A» ^ + 3 A»A -^ + 3 AA* -^. + A» ?^, 

^ ' aa^ aar'ay aj«/ ay 

and so on. 

Substituting these results in {E), we get 

(«4) /(x + M, y + ^) =/(aB, ») + «(* ^ + * ^) 

To get /(a: + A, y + A), replace t by 1 in (M), giving Taylor's 
Theorem for a function of two independent variables, 

(«5) f{!c + h,v + k)=:f{x,v) + h^ + k^ 

+ ;^U,^^ + 2Hk^ + k'm+..., 
|2\ cte« aasay Oy*/ 

which is the required expansion in powers of h and k. Evidently 
(65) is also adapted to the expansion of f{x -j- A, y -j- A:) in powers 
of X and y by simply interchanging x with h and y with i. Thus, 

(«5o) f(x + h,y + k)=f(h,k) + x^^ + y^ 



+^K^+''^^+»'^)+ 



|2\ a;k« • '^ aMit ' " dk* 
Similarly for three variables we shall find 
(66) f{x + h,y + k,z + l)=f(x,y,z) + h^ + k^ + l^( 



^I2V c>x"^ c)f/»^ c)«>^ 



[2V 5x' c)y» dz* dxdy 

' d;edx dydz ' 

and so on for any number of variables. 



246 DIFFERENTIAL CALCULUS 



EXAMPLES 

1. Given /(x, y) = Ax^ + Bxy + Cy" ; expand /(x + A, y + *) In powera of h 
and ib. 

Solution, — = 2Ax + By, — =JBx + 2Cy; 

dx ^ 

dx^ dxdy ay« 

The third and higher partial deriyatives are all zero. Sabstituting in (65), 

/(x + A, 1/ + *) = -4x2 + Bxy + Cj/» + (2 -4x + By) A + (BiB + 2 Cy)* 

+ Ah^ + B^iifc + Cifc». An*. 

2. Given /(x, y, «) = ^x« + By^ + Cz^ ; expand /(x + i, y + m, « + n) in powerg 
of {| m, n. 

5o/tt<ion. — = 2^x, — = 2By,^ = 2Cz; 

dx ' dy az 

ex2 c>y2 az« axay eyaz dzdx 

The third and higher partial derivatives are all zero. Substituting in (66), 

/(x + ^ y + rn, z-f n) = -4x« + 5y2+ Cz« + 2 -4xZ + 2 Bym + 2 Czn 

+ -4Z2 + Bma + Cn^, Ana. 

3. Given /(x, y) = Vx tan y ; expand/(x + A, y + ^) in powers of h and k, 

4. Given /(x, y, z) = -4xa -f By« + Cz^ + Dxy + ^yz + l^zx ; expand /(x + A, 
y -^ kf z + in powers of h, k^ I. 

162. Maxima and minima of functions of two independent vari- 
ables. The function /(2:, y) is said to be a maximum at 2; = a, y = i 
when /(a, b) is g^reater than f{x^ y) for all values of x and y in the 
neighborhood of a and b. Similarly /(a, b) is said to be a minimum 
at 2; = a, y = i when /(a, 6) is less than f(x^ y) for all values of x 
and y in the neighborhood of a and &. 

These definitions may be stated in analytical form as follows : 
If, for all values of h and k numerically less than some small 
positive quantity, 

(A) f{a + A, b-^k) —f{a^ 6) = a negative number^ then /(a, b) is 
a maximum value of /(a:, y). If 

(5) f{a -t- A, 6 -f- A:) —f(a^ b)=:a positive number^ then /(a, 6) is 
a minimum value of /(x, y). 



EXPANSION OF FUNCTIONS 



247 




These statements may be interpreted geometrically as follows: 
a point P on the surface ^. . 

is a maximum point when it is ^^ higher" than all other points on 
the surface in its neighborhood, the coordinate plane XOY being 
assumed horizontal. Similarly P' is a minimum point on the 
surface when it is 
«* lower " than all 
other points on. the 
surface in its neigh- 
borhood. It is there- 
fore evident that 
all vertical planes 
through P cut the 
surface in curves (as 
AFB or DFE in the 
figure), each of which 
has a maximum ordinate z {^MP) at P. In the same manner 
all vertical planes through P' cut the surface in curves (as BP^C 
or FP^O)y each of which has a minimum ordinate z {z^NP^) at P'. 
Also, any contour (as HIJK) cut out of the surface by a horizontal 
plane in the immediate neighborhood of P must be a small closed 
curve. Similarly we have the contour LSBT near the minimum 
point P'. 

It was shown in §§ 93, 94, pp. 117-121, that a necessary con- 
dition that a function of one variable should have a maximum 
or a minimum for a given value of the variable was that its first 
derivative should be zero for the given value of the variable. 
Similarly for a function f(x^ y) of two independent variables, a 
necessary condition that /(a, b) should be a maximum or a minimum 
(i.e. a turning value) is that for a: = a, y = 6, 

^ = 0, ^ = 0. 

dx dy 

Proof. Evidently {A) and (P) must hold when A = ; that is, 

f{a-\.Kh)^f{a,l) 
is always negative or always positive for all values of h sufficiently 
small numerically. By §§ 93, 94, a necessary condition for this is 



{C) 



248 DIFFERENTIAL CALCULUS 

that -r-f(x^ b) shall vanish for a: = a, or, what amounts to the same 
ax 

thing, '^rf^'i V) ^^ vanish for a: = or, y = 6. Similarly {A) and 

ox 

(B) must hold when A = 0, giving as a second necessary condition 
that — f{Xy y) shall vanish for a; = a, y = ft. 

In order to determine sufficient conditions that /(a, ft) shall be 
a maximum or a minimum it is necessary to proceed to higher 
derivatives. To derive sufficient conditions for all cases is beyond 
the scope of this book.* The following discussion, however, will 
suffice for all the problems given here. 

Expanding f(a -j- A, ft -h A:) by Taylor's Theorem, (65), p. 245, 
replacing a; by a and y by ft, we get 

ox cy 



-|("S--S-*-|)--. 



where the partial derivatives are evaluated for a; = a, y = ft, and R 
denotes the sum of all the terms not written down. All such 
terms are of a degree higher than the second in A and k. 

Since — = and r^ = 0, from (C), p. 247, we get, after transpos- 
dx oy 

ing/(a, ft), 

(^) fia + k^h + k) -fia, h) = 1 (A«g -f 2M^ + *»!) + B. 

If /(a, ft) is a turning value, the expression on the left-hand 
side of (U) must retain the same sign for all values of A and k suffi- 
ciently small in numerical value, the negative sign for a maximum 
value [(J.), p. 246] and the positive sign for a minimum value 
[(S), p. 246], i.e. /(a, ft) will be a maximum or a minimum accord- 
ing as the right-hand side of (U) is negative or positive. Now E 
is of a degree higher than the second in A and k. Hence as A and k 
diminish in numerical value it seems plausible to conclude that the 
numerical value of B will eventually become and remain less than 

• See Courg d 'Analyact Vol. I, by CL Jordan. 



EXPANSION OF FUNCTIONS 249 

the numerical value of the sum of the three terms of the second degree 
written doum on the right-hand side of (E),* Then the sign of the 
right-hand side (and therefore also of the left-hand side) will be 
the same as the sign of the expression 

OTT. dxdy dy 

But from Algebra we know that the quadratic expression 

always has the same sign as A {or B) when AB — C^> 0. 

Applying this to (F\ A = —,^ B = -^, C = — =^» and we see that 
^^•^ ^ ^ ^ dJ^ df dxdy 

{F)j and therefore also the left-hand member of (H), has the same 



sign as g(or^) when 



ay ^ _ /_^Y^ 

cj^ dt/^ \dxdyj 

Hence the following rule for finding maximum and minimum 
values of a function /(;r^ y). 

First step. Solve the simultaneous equations 

^ = 0,^ = 0. 

dx dy 

Second step. Calculate for these values of x and y the value of 

gygy /gyy 

dJ? df \dxdyj ' 
Third step. The function vrill have a 

maximum ?/ A > and ;— f or t^ J < 0; 

minimum if A>0 and ^ ^or ^)>0; 

neither a maximum nor a minimum ij A < 0. 
The question is undecided i/ A = O.f 

•Peano has shown that this conclusion does not always hold. See the article on *< Maxima 
and Minima of Functions of Several Variables," by Professor James Pierpont in the BiUletin 
of the American Mathematical Society ^ Vol. IV. 

t The discussion of the text merely renders the given rule plausible. The student should 
observe that the case A = is omitted in the discussion. 



250 DIFFERENTIAL CALCULUS 

The student should notice that this rule does not necessarily 
give all maximum and minimum values. For a pair of values 
of X and y determined by the first step may cause A to vanish, 
and may lead to a maximum or a minimum or neither. Further 
investigation is therefore necessary for such values. The rule is, 
however, sufficient for solving many important examples. 

The question of maxima and minima of functions of three 
or more independent variables must be left to more advanced 
treatises. 

Ex. 1. Examine the fonction Saxy — x* — y^ for maximam and minimum 
yalues. 

Solutum, f{z, y) = 8 axy — x* — y*. 

First step. ^ = 3ay - Sx* = 0, — = 3aa; -3y« = 0. 

bx dy 

Solving these two eqaations simultaneously, we get 

X = 0, X = a, 

y = 0; y = a. 

Second step. -4 = -6x, — ^ = 3a, -4=-6y; 

dz^ dxdy ey« '^ 



5x« dy^ \dzdv/ 



dy^ \dxdy 

Third step. When x = and y = 0, A = — a^, and there can be neither a 
maximum nor a minimum at (0, 0).. 

ay 

When X = a and y^^a, A = + 27 a^ ; and since -^ = — 6 a, we haye the condi- 
tions for a maximum yalue of the function fulfilled at (a, a). Substituting x = a, 
y = a in the giyen function, we get its maximum yalue equal to a*. 

Ex. 2. Divide a into three parts such that their product shall be a maximum. 

Solution, Let x = first part, y = second part ; then a — {z-\-y)z:::a^x-'y = 
third part, and the function to be examined is 

/(x, y) = xy{a-x-'y). 

First step. -^=zay - 2xy - y^ - o, — = ox - 2xy - x« = 0. 
dx By 

Solving simultaneously, we get as one pair of values x = - « y = -.* 

3 8 

Secondstep. -^ = - 2y, — ^ = a — 2x — 2y, — = — 2x: 
axa dxdy dy^ 

A = 4xy - (a - 2x - 2y)2. 

** a?" 0, y » are not considered, since from the nature of the problem we would then have a 
minimum. 



EXPANSION OF FUNCTIONS 251 

Third step. When a; = - and y = -» A = — : and since — = — --, it is seen 

3 8 3 ' 8x« 3 

that our product is a maximum when x = - > y = -. Therefore the third part is 

also - , and the maximum value of the product is — • 
3 27 



EXAMPLES 

1. Find the maximum value oix^ + xy -^ j/* — ax — by. Ans. i (oft — a* — 6"). 

3ir 

2. Show that sin 2 + sin y 4- cos (x 4- y) is a minimum when x = y = — * and 

tr ^ 

a maximum when x = y = — • 

3. Show that x&f-^'^y has neither a maximimi nor a minimum. 

lax A-hu A- c\^ 

4. Show that the maximum value of ^ „ * is a« + fts + c^. 

x3 + ya + 1 

6. Find the greatest rectangular parallelopiped that can be inscribed in an 
ellipsoid. That is, find the maximum value of 8 xyz(= volume) subject to the 
condition a-a y* «« ^ %aJbc 

a« 62 c2 3V8 

HitU. Let ic >■ xys, and substitute the value of « ftom the equation of the ellipsoid. This gives 

where « is a function of only tvo variables. 

6. Show that the surface of a rectangular purallelopiped of given volume is 
least when the solid is a cube. 

7. Examine x* 4- y* — 2^ 4- xy -> y> f or maximum and minimum values. 

Ara. Maximum when x = 0, y = ; 

minimum when x = y = ± i, and when x = — y = i: i V3. 

8. Show that when the radius of the base equals the depth, a steel cylin- 
drical standpipe of a given capacity requires the least amount of material in its 
construction. 

9. Show that the most economical dimensions for a rectangular tank to hold a 
given volume are a square base and a depth equal to one half the side of the base. 

10. The electilc time constant of a cylindrical coil of wire is 

mxyz 

ax-\-oy -\- cz 

where x is the mean radius, y is the difference between the internal and external 

radii, z is the axial length, and m, a, 6, c are known constants! The volume of 

the coil is fixyz = g. Find the values of x, y, z which make u a minimum if the 

volume of the coil is fixed. zlabcg 

Am. ax=zby^cz=\^ ' 



CHAPTER XXI 



ASYMPTOTES. SINGULAR POINTS. CURVE TRACING 




163. Rectilinear asymptotes. An asymptote to a curve is the 
limiting position* of a tangent whose point of contact moves off 
to an infinite distance from the origin.f 

Thus, in the hyperbola, the asymptote AB is the limiting 
position of the tangent FT as the point of contact F moves off 

to the right to an infinite distance. In 
the case of algebraic curves the follow- 
ing definition is useful : an asymptote is 
the limiting position of a secant as two 
points of intersection of the secant with a 
branch of the curve move off in the same 
direction along that branch to an infinite 
distance. For example, the asymptote 
AB is the limiting position of the secant 
FQ as F and Q move upwards to an infinite distance. 

164. Asymptotes found by method of limiting intercepts. The 
equation of the tangent to a curve at {x^^ y^ is by (1), p. 89, 

First placing y = and solving for a:, and then placing a;= 
and solving for y, and denoting the intercepts by x^ and y, 

respectively, we get 

dx 
Xi = a:r^ — y^ - ^ = intercept an OX; 

y^=y^ — Xj -7^ = intercept on OY. 

* A line that approaches a fixed straight line as a limitinp; position cannot be wholly at 
infinity ; hence it follows that an asymptote must pass within a finite distance of the origin. 
It is evident that a curve whicli has no infinite brancli can have no real asymptote. 

t Or, less precisely, an asymptote to a curve is sometimes defined as a tangent whoee point of 
contact is at an infinite distance. 

252 



ASYMPTOTES 253 

Since an asymptote must pass within a finite distance of the 

origin, one or both of these intercepts must approach finite values 

as limits when the point of contact {x^, y^ moves off to an infinite 

distance. If 

limit (a-,) = a and limit (y,) = 6, 

then the equation of the asymptote is found by substituting the 
limiting values a and b in the equation 

a h 
If only one of these limits exists, but 

then we have one intercept and the slope given, so that the 
equation of the asymptote is 

y = mx-\-b or a: = — -f a. 

Ex. 1. Find the asymptotes to the hyperbola — = 1. 

« , ^. dy K^x 6 1 , limit /dy\ h 

SoltUion. — = — = ± , and 7»= """^ ( -^ ) = ±-. 

dx a^y a I ^ ^ = QoVdx/ a 

Also, a^ = — and y. = ; hence these intercepts are zero when x = y = ao. 

Therefore the asymptotes pass through the origin (see figure on p. 252) and their 
eqoations are 

y — = ± - (x — 0), or, ay = ±bx. Ana. 

vw 

This method is frequently too complicated to be of practical 
use. The most convenient method of determining the asymptotes 
to algebraic curves is given in the next section. 

165. Method for determining asymptotes to algebraic curves. 
Given the algebraic equation in two variables, 

(A) fix, y) = 0. 

If this equation when cleared of fractions and radicals is of 
degree n, then it may be arranged according to descending powers 



254 DIFFERENTIAL CALCULUS 

m 

of one of the variables, say y, in the form 

For a given value of a^ this equation determines in general n 
values of y. 

Case I. To determine the asymptotes to the curve (B) which are 
parallel to the coordinate axis. Let us first investigate for asymp- 
totes parallel to OF. The equation of any such asymptote is of 
the form 

(C) x = k, 

and it must have two points of intersection with (B) having infinite 
ordinates. 

First, Suppose a is not zero in (£), that is, the term in y" is 
present. Then for any finite value of 2:, (B) gives n values of y, 
all finite. Hence all such lines as ((7) wiU intersect (B) in points 
having finite ordinates, and there are no asymptotes parallel to OY, 

Second, Next suppose a = but ( and c are not zero. Then we 
know from Algebra that one root (= y) of (B) is infinite for every 
finite value of x; that is, any arbitrary line (O) intersects (B) at 
only one point having an infinite ordinate. If now in addition 

6a: -f <? = 0, or, 

then the first two terms in (B) will drop out, and hence two of its 
roots are infinite. That is, (2)) and (B) intersect in two points 
having infinite ordinates, and therefore (J9) is the eqtuUion of an 
asymptote to (B) which is parallel to OY. 

Third. If a = 6 = <? = 0, there are two values of x that make y 
in {B) infinite, namely, those satisfying the equation 

{E) dar»-f ex-f/=0. 

Solving (E) for x, we get two asymptotes parallel to OYj and so 
on in general. 

* For ufle in this section the attention of the student Is called to the following theorem from 
Algebra : Given an algebraic equation of degree n, 

^y" + J5y" -» + Cy" ->+ JDy»- « + •••= 0. 

When A approaches zero, one root (value of y) approaches «. 

When A and B approach zero, two roots approach oo. 

When Af B, and C approach zero, three roots approach oo, etc. 



ASYMPTOTES 



255 



In the same way, by arranging /(a:, y) according to descending 
powers of x^ we may find the asymptotes parallel to OX. Hence 
the following rule for finding the assrmptotes parallel to the coor- 
dinate axes : 

First step. Equate to zero the coefficient of the highest power of x 
in the equation. This gives all asymptotes parallel to OX, 

Second step. Etpiate to zero the coefficient of the highest power of 
y in the eqaatum. This gives all asymptotes parallel to OY. 

Note. Of course if one or both of these coefficients do not 
involve x (or y), they cannot be zero, and there will be no corre- 
sponding asymptote. 

Ex. 1. Find the asymptotes of the carre aH = y (x -^ a)*. 

Solution. Arranging the terms according to powers of z, 

yx« - (2 ay + a^a; + a^ = 0. 

Equating to zero the coefficient of the highest power of x, we get y = as the. 
asymptote parallel to OX. In fact the 
asymptote coincides with the axis of x. 
Arranging the terms according to powers 

^^^' (x-a)«y-a«x =0. 

Placing the coefficient of y equal to 
zero, we get X = a twice, showing that 
^B is a double asymptote parallel to OY. 
If this curve is examined for asymptotes 
oblique to the axes by the method explained 
below, it will be seen that there are none. Hence y = and x = a are the only 
asymptotes of the given curve. 

Case II. To determine asymptotes oblique to the coordinate axes. 

Given the algebraic equation 

(F) /(^y)=o. 

Consider the straight line 

(Q) y=zmx-\-k. 

It is required to determine m and k so that the line ( 69^) shall be 
an asymptote to the curve (F), 

Since an asymptote is the limiting position of a secant as two 
points of intersection on the same branch of the curve move off to 
an infinite distance, if we eliminate y between (F) and ((?), the 
resulting equation in rr, namely, 

(H) f{x,mx-^h)^Q, 




256 



DIFFERENTIAL CALCULUS 



must have two infinite roots. But this requires that the coeffi- 
cients of the two highest powers of x shall vanish. Equating these 
coefficients to zero, we get two equations from which the required 
values of m and h may be determined. Substituting these values 
in ((?) gives the equation of an asymptote. Hence the following 
rule for finding asymptotes oblique to the coordinate axes : 

First step. Replace y by mx-\-k in the given equation and expand. 

Second step. Arrange the terms according to descending powers 
of X. 

Third step. Equate to zero the coefficients of the two highest 
powers* of Xy and solve for m and Ic, 

Fourth step. Substitute these values of m and k in 

y = mx -f k. 
This gives the required asymptotes. 

Ex. 2. Examine ^ = 2 ox^ - x^ for asymptotes. 

Solution, Since none of the terms involve both x and y, it is evident that there 
are no susymptotes parallel to the coordinate axes. To find the oblique asymptotes^ 

eliminate y between the given equation and 
y = mx -\- k. This gives 

(?nx + A:)« = 2ax«-z»; 

and arranging the terms in powers of x, 

(1 + m8)x' + (3 m«ik - 2 a)x2 + 3 k^mx + *» = 0. 

Placing the first two coefficients equal to zero, 

1 + m* = and 3 rn-^k - 2 a = 0. 

2a 




Solving, we get ?7i = — 1 , A: = 



Substituting 



2a 



in y = 7HX + fc, we have y=z—x-\ , the equa- 
tion of asymptote AB. 



EXAMPLES 

Examine the first eight curves for asymptotes by the method of § 164, and the 
remaining ones by the method of § 165. 



1. y = e*. 



Ans. y = 0. 



2. y 



Arts, y = 0. 



* If the term involving t"~H8 missing, or if tbe value of m obtained by placing the first 
coefficient equal to zero causes the second coefficient to vanish, then by placing the ooefflcieuts 
of x'* and x^~2 equal to zero we obtain two equations from which the values of m and k may be 
found. In this case we shall in general obtain two k*s for each m, that Is, pairs of parallel oblique 
asymptotes. Similarly, if the term in x**-^ is also missing, each value of m furnishes three 
parallel oblique asymptotes, and so on. 



ASYMPTOTES 



257 



3. y = logx. 



1\* 



6. y =z tan x. 
1 



An8, x = 0. 

n being any odd integer, x 

6. y = c* - 1. X = 0, y = 0. 

7. y» = 6x« + x». y = x-|-2. 

8. Show that the parabola has no a^mptotes. 

9. y* = a' — x». y + X = 0. 



UT 



10. The cisBoid y^ = 

11. y«a = yax + x». 



x« 



2r-x 



12. y«(xa + l) = x«(x»-l). 

13. y2(x~2a) = x«-o». 

14. xV = a«(x? + y«). 

15. y(z*-36x + 26«) = x'-3ax« + a«. 

16. y = c4- 



o« 



x = 


2r. 




* 


X = 


a. 






y = 


:iX. 






x = 


2a, y 


= ± (X + a). 




x = 


±a, » 


= ± a. 




x = 


6, x = 


26, y + 3a = 


X + 3&. 


y = 


c, x = 


b. 




y + 


X + a : 


= 0. - 




y = 


0. 






x = 


0, y = 


0, X + y = 0. 





(X - 6)» 

17. The folium x» + y« - 3axy = 0. 

18. The wltchx^y = 4a«(2a-y). 

19. xi/<^ -\- x"^ := a\ 

20. x» + 2x«y-xy*-2y» + 4y« + 2xy + y = 1. 

x + 2y = 0, x + y = l, x-y = -l. 

166. Asymptotes in polar coordinates. Let /(/>, 0) = be the 
equation of the curve FQ having the asymptote CD. As the 
asymptote must pass within a finite 
distance (as OU) of the origin^ and the 
point of contact is at an infinite dis- 
tance, it is evident that the radius 
vector Oi^ drawn to the point of contact 
is parallel to the asymptote, and the 
subtangent OE is perpendicular to it. Or, more precisely, the 
distance of the asymptote from the origin is the limiting value of 
the polar subtangent as the point of contact moves off an infinite 
"distance. 




268 DIFFERENTIAL CALCULUS 

To determine the asymptotes to a polar curve, proceed as follows : 
First step. Find from the equation of the curve the values of 6 

which make p — oo.* These values of give the directions of the 

asymptotes. 

Second step. Find the limit of the polar subtangent 

as 6 approaches each such value^ remembering that p approaches oo 
at the same time. 

Third step. If the limiting value of the polar suhtangent is finite^ 
there is a corresponding asymptote at that distance from the origin 
and parallel to the radius vector drawn to the point of contact. 
When this limit is positive the asymptote is to the rights and when 
negative^ to *he left of the origin^ looking in the direction of the infinite 
radius vector. 

EXAMPLES 

1. Examine the hyperbolic spiral p = - for asymptotes. 

V 

SohUion, When ^ = 0, p = oo. Also — = — — , hence 

d$ e» 

d0 a^ 0^ 
Subtangent = p2— = — . = — <L 

* *^ dp 0^ a 




^= ^~] = " ^' ^^^^^ ^ ^^*®- 



It happens in this case that the subtangent is the 
same for all values of e. The curve has therefore an 
asymptote BC parallel to the initial line OA and at a distance a above it. 

Examine the following curves for asymptotes. 

2. pcos9 = acos2^. 

Ans. There is an asymptote perpendicular to the initial line at a distance a 
to the left of the origin. 

3. p = a tan B, 

AnA. There are two asymptotes perpendicular to the initial line and at a dis- 
tance a from the origin, on either side of it. 

4. The lituus p^ = a. Am, The initial line. 

* If the equation can be written as a polynomial in p, these Talnes of 9 may be found I9 
equating to zero the coefficieDt of the highest power of p (see footnote, p. 254). 



SINGULAR POINTS 259 

6. p = a sec 2 ^. 

Ans. There are four asymptotes at the same distance - from the origin, and 
inclined 45'' to the initial line. ^ 

6. (p ~ a) sin tf = 6. 

Ana, There is an asymptote parallel to the initial line at the distance b 
above it. 

7. P=a (sec 2 ^ + tan 2 0). 

An9, There are two asymptotes parallel to 9 = -» at the distance a on each 
side of the origin. 

8. Show that the initial line is an asymptote to two branches of the curve 
^sin^ = a3cos2^. 

9. Parabola p ?— . An,. There is no asymptote. 

1-C08« 

167. Singular points. Given a curve whose equation is 

f{x,t,) = 0. 
Any point on tlie curve for which 

^ = Oand^ = 
dx dy 

is called a %ingular point of the curve. All other points are called 
ordinary points of the curve. Since by {55 a), p. 202, we have 

dy _ dx 

dy 
it is evident that at a singular point the direction of the curve 
(or tangent) is indeterminate, for the slope takes the form -• In 

the next section it will be shown how tangents at such points may 

be found. 

168. Determination of the tangent to an algebraic curve at a given 
point by inspection. If we transform the given equation to a new 
set of parallel coordinate axes having as origin the point in ques- 
tion on the curve, we know that the new equation will have no 
constant term. Hence it may be written in the form 

{A) f{x, y)=ax + by-{-{ex^-{- dry -f ef) 

-\-{f^'\-g^y +^^f + «y) + ••• = 0, 



260 DIFFERENTIAL CALCULUS 

The equation of a tangent to the curve at the given point (now 
the origin) will be 

(5) y = @)^- By (1), p. 89 

Let y^mx (by 54 (c), p. 3) be the equation of a line through 
the origin and a second point F on the locus of {A). If then F 
approaches along the curve, we have from {B) 

(0) limit w = $^. 

ax 

Let be an ordinary point. Then, by § 167, a and b do not 
both vanish since at (0, 0), from (A), 

dx dy 

Replace y in (A) by mx^ divide out the factor x^ and let x 
approach zero as a limit. Then (A) will become* 

a 4- Am = 0. 

Hence we have from (-B) and (0) 

aa; -f 6y = 0, 

the equation of the tangent. The left-hand member is seen to 
consist of the terms of the first de^ee in {A). 

When is not an ordinary point we have a = 6 = 0. Assume 
that c, (2, e do not all vanish. Then proceeding as before (except 
that we divide out the factor x'), we find, after letting x approach 
the limit zero, that {A) becomes 

c -f dm -f eni? = 0, 
or, from (0), 

<^ -<f)-(i)'-»- 

* After dividing by x an algebraio equation In m remains whose coefficients are f nnotlons of x. 
If now X approaches zero as a limit, the theorem holds that one root of this equation in m will 
approach the limit — o-r 6. 



SINGULAR POINTS 261 

Substituting from (£), we see that 

(H) C2? -f dxy -f e^ = 

IS the equation of the pair of tangents at the origin. The left- 
hand member is seen to consist of the terms of the second degree 
in (A). Such a singular point of the curve is called a double point 
from the fact that there are two tangents to the curve at that 
point. 

Since at (0, 0), from (-4), 

d3? dxdy df ' 

it is evident that {D) may be written in the form 

^^ dj?^ dxdy\dx)^df\dx) 

In the same manner, if 

there is a triple point at the origin, the equation of the three tan- 
gents being 

fj? -f g2?y -f hxy"" -{-iy'' = 0. 

And so on in general. 

If we wish to investigate the appearance of a curve at a given 
point, it is of fundamental importance to solve the tangent prob- 
lem for that point. The above results indicate that this can be 
done by simple inspection after we have transformed the origin to 
that point. 

Hence we have the following rule for finding the tangents at a 
given point. 

First step. Transform the origin to the point in question. 

Second step. Arrange the terms of the resulting equation accord- 
ing to ascending powers of x and y. 

Third step. Set the group of terms of lowest degree equal to zero. 
This gives the equation of the tangents at the point {origin). 



262 



DIFFERENTIAL CALCULUS 




Ex. 1. Find the eqaation of the tangent to the 

^^ 6x« + 5y2 + 2xy-12aj-12y = 
at the origin. 

Solution. Placing the terms of lowest (first) degree 
equal to zero, we get 

-12x-12y = 0, 
or, X + y = 0, 

which is then the equation of the tangent PT at the 
origin. 

Ex. 2. Examine the curre 3 x^ — xy — 2 y^ + £' — 8 y* = f or tangents at the 
origin. 

SoliUion. Placing the terms of lowest (second) 
degree equal to zero, 

3x«-xy- 2 2^ = 0, 

or, (x-y)(8x + 2y) = 0, 

X — y = being the eqiiation of the tangent AB, 
and 3 X + 2 y = the equation of the tangent CD. 
The origin is, then, a double point of the curve. 

Since the roots of the quadratic equation (F), p. 261, namely, 

df \dx) dxdy \dx) d3? 

may be real and unequal, real and equal, or imaginary, there are 
three cases of double points to be considered, according as 

^ ^ \dxdy) d^dy" 

is positive, zero, or negative (see 3, p. 1). 

169. Nodes. 




( 



\dxdy) ax* dy^ 
In this case there are two real and unequal values of the slope 

= -T-) found from (jP), so that we have two distinct real tangents 

to the curve at the singular point in question. This means that 
the curve passes through the point in two different directions, or, 
in other words, two branches of the curve cross at this point. 
Such a singular point we call a real double point of the curve, or a 
node. Hence the conditions to be satisfied at a node are 

fee i/)-o ^~o ^-o (J!L.y^^f^f 



dx^dy^ 



>0. 




SINGULAR POINTS 268 

Ex. 1. Examine the lemniscate y* = x^ — x* for singular points. 
Solution. Here /(x, y) = y*--x*4.x* = 0. 

Also, ^ = -2x + 4x8 = 0, ^ = 2y = 0. 

dx by 

The point (0, 0) is a singular point since its coordinates satisfy the above three 
equations. We have at (0, 0), 

g = -2. -^=0,^ = 2. 
ax* bxby ay« 

\bxdy/ dx^dy^ ' 

A' 

and the origin is a double point (node) through which 

two branches of the curve pass in different directions. By placing the terms of the 

lowest (second) degree equal to zero we get 

y2 — xa = 0, or y = X and y = — x, 

the equations of the two tangents AB and CD at the singular point or node (0, 0). 

na Cusps. (j?!Ly_^^^ = o. 

In this case there are two real and equal values of the slope 
found from (-F), hence there are two coincident tangents. This 
means that the two branches of the curve which pass through the 
point are tangent. When the cui-ve recedes from the tangent in 
both directions from the point of tangency, the singular point is 
called a point of oscillation ; if it recedes from the point of tangency 
in one direction only, it is called a ciisp. There are two kinds 
of cusps. 

First kind. When the two branches lie on opposite sides of 
the common tangent 

Second kind. When the two branches lie on the same side of 
the common tangent.* 

The following examples illustrate how we may determine the 
nature of singular points coming under this head. 

Ex. 1. Examine a*j/^ = a%c* — sfi for singular points. 
Solution. Here /(x, y) = aV - a^x* + x« = 0, 

^ = -4a^8 + 6x» = 0, ^ = 2a*y = 0, 
dx dy 

• Meaning in the neighborhood of the Binguliir point. 



264 



DIFFERENTIAL CALCULuS 



and (0, 0) is a singular point since it satisfies the above tliree equations. Also, at 

(0, 0) we have 

ay 



7} 



e*^ 



ax« dxdy 



^■^=0.^ = 20.. 
ey2 






and since the curve is symmetrical with respect to OF, 
the origin is a point of osculation. Placing the terms of 

lowest (second) degree equal to zero, we get y^ = 0, showing that the two common 

tangents coincide with OX. 



Ex. 2. Examine ^ = x* for singular points. 
Solution. Here /(x, y) = y^ - x» = 0, 



^ = -3x» = 0, ^ = 2y = 0, 
dx dy 



showing that (0, 0) is a singular point Also, at (0, 0) we have 

\bxdyJ cx^dy^ 



ex* ' (XOy 



= o.|?C = 2. ../^:^y-^.^ = o. 



ays 




This is not a point of osculation, however, for if we solve the 
given equation for y, we get 

y=iVx8, 

which shows that the curve extends to the right only of OT^ for negative values of 
X make y imaginary. The origin is therefore a cusp, and since the branches lie 
on opposite sides of the common tangent it is a cusp of the first kind. Placing the 
terms of lowest (second) degree equal to zero, we get y^ = 0, showing that the two 
common tangents coincide with OX. 

Ex. 3. Examine (y — x^)^ = x' for singular points. 

SolMiUm. Proceeding as in the last example, we 
find a cusp at (0, 0), the common tangents to the two 
branches coinciding with OX. Solving for y, 

y = X* ± x'. 

If we let X take on any value between and 1, 
y takes on two different positive values, showing that 
in the vicinity of the origin both branches lie above 
the common tangent. Hence the singular point (0, 0) 
is a cusp of the second kind. 

171. Conjugate or isolated points, f ^^ V- ^ ^ < o. 

In this case the values of the slope found from (B) are imagi- 
nary. Hence there are no real tangents; the singular point is 
the real intersection of imaginary branches of the curve, and the 




SINGULAR POINTS 



265 



coordinates of no other real point in the immediate vicinity satisfy 
the equation of the curve. Such an isolated point is called a 
conjugate point, 

Ex. 1. Examine the carve |/^ = x* — £> for singular points. 
Solution. Here (0, 0) is found to be a singular point of the curve 

/hi 

at which ~ = ± V— 1. Hence the origin is a conjugate point. Solv- 
dx 

ing the equation for y, 



O 



T 



y = ± aj Vx-1. 

This shows clearly that the origin is an isolated point of the curve, 
for no values of x between and 1 give real values of y. 



172. Transcendental singularities. A curve whose equation in- 
volves transcendental functions is called a transcendental curve. 
Such a curve may have an end pointy at which it terminates abruptly, 
caused by a discontinuity in the function ; or a salient point at 
which two branches of the curve terminate without having a 
common tangent, caused by a discontinuity in the derivative. 



Ex. 1. Show that ^ = x log x has an end point at 
the origin. 

Solution. X cannot be negative since negative num- 
bers have no logarithms; hence the curve extends only 
to the right of OT, When x = 0, y = 0. There being 
only one value of y for each positive value of x, the curve 
consists of a single branch terminating at the origin, 
which is therefore an end point. 




Ex. 2. Show that y = 



j- has a salient point at the origin. 

1 + c* 



Solution. Here — = 

dx 



7 + 



1 




1 + c* x(l + ci)« 

If X is positive and approaches zero as a limit, 
we have ultimately 

y = and -^ = 0. 
dx 

If X is negative and approaches zero as a limit, we get ultimately 

dy 
2/ = and -- = 1. 
dx 

Hence at the origin two branches meet, one having OX as its tangent and the 
other, ABj making an angle of 46*^ with OX. 



266 DIFFERENTIAL CALCULUS 



EXAMPLES 

1. Show that ^ = 2 xs + X* has a node at the origin, the slopes of the tangents 
being db V5. 

2. Show that the origin is a node of ^(a^ + x*) = x»(a» - x>), and that the 
tangents bisect the angles between the axes. 

3. Prove that (a, 0) is a node of y^ = x (x — a)', and that the slopes of the 
tangents are db^^* 

4. Prove that a^^ — 2 aJbz^ —ofi^O has a point of osculation at the origin. 

6. Show that the curve ^ = x^ + x* has a point of osculation at the origin. 

x' 

6. Show that the cissoid y^ — has a cusp of the first kind at the origin. 

2a — X 

7. Show that ^ = 2 ox^ — x' has a cusp of the first kind at the origin. 

8. In the curve (y — x^Y = ^^ show that the origin is a cusp of the first or 
second kind according as n is < or > 4. 

9. Prove that the curve x* — 2 ax*y — oxy* + cfiy'^ = has a cusp of the second 
kind at the origin. 

10. Show that the origin is a conjugate point on the curve y^ (x^ — a^ = x*. 

11. Show that the curve y> = x (a + x)^ has a conjugate point at (— a, 0). 

12. Show that the origin is a conjugate point on the curve ay^ — x* + bx^ = 
when a and h have the same sign, and a node when they have opposite signs. 

13. Show that the curve x* + 2 ox^ — ay* = has a triple point at the origin, 
and that the slopes of the tangents are 0, + V^, and — v^. 

14. Show that the points of intersection of the curve (-) +(-) =1 with the 
axes are cusps of the first kind. ^ 

15. Show that no curve of the second or third degree in x and y can have a cusp 

of the second kind. 

_i 

16. Show that y = e ' has an end point at the origin. 

17. Show that y = x arc tan - has a salient point at the origin, the slopes of the 

T * 

tangents being ± - • 

173. Curve tracing. The elementary method of tracing (or 
plotting) a curve whose equation is given in rectangular coordi- 
nates, and one with which the student is already familiar, is to 
solve its equation for y (or a:), assume arbitrary values of x (or y), 
calculate the corresponding values of y (or x), plot the respective 



CURVE TRACING 267 

points, and draw a smooth curve through them, the result being 
an approximation to the required curve. This process is laborious 
at best, and in case the equation of the curve is of a degree higher 
than the second, the solved form of such an equation may be unsuit- 
able for the purpose of computation, or else it may fail altogether, 
since it is not always possible to solve the equation for y or x. 

The general form of a curve is usually all that is desired, and 
very often we care to examine the curve in the neighborhood of a 
certain point only. To attain this object it is as a rule only neces- 
sary to determine some of the important points, lines, and proper- 
ties of the curve as enumerated below. 

No rules for tracing a curve can be given that will apply in all 
cases, but the student will find it to his advantage to use the fol- 
lowing general directions as a guide and to study carefully the 
examples that are worked out in detail. 

174. General directions for tracing a curve whose equation is given 
in rectangular coordinates. 

1. Examine the curve for symmetry. 

(a) If the equation is unchanged when y is replaced by — y, 
the curve is symmetrical with respect to OX, 

(b) If the equation is unchanged when x is replaced by — a:, 
the curve is symmetrical with respect to F. 

(c) If the equation is unchanged when x is replaced by — a:, 
and y by — y, the curve is symmetrical with respect to the origin 
which is also the center of the curve. 

2. Examine the curve for important points, 

(d) If the equation is satisfied by a; = 0, y = 0, the curve 
passes through the origin. 

(e) Placing x = and solving for y gives the intercepts on 
OY. Placing y = and solving for x gives the intercepts on OX, 

(f) Find -^; this gives the direction of the curve at any point 
and serves to locate maximum and minimum points (§ 94, p. 120). 

(g) Find -7^; this gives the direction of curvature at aP" 

point and serves to find the points of inflection (§ 98, p. 137 
(h) Examine the curve for singular points (p. 259). 



268 



DIFFERENTIAL CALCULUS 



3. (i) Uxamine the curve for OBymptotes (§§ 164, 165, p. 252). 
Determine on which side of each asymptote the corresponding 

infinite branch lies. 

4. (j) Locate additional points on the curve. If possible, com- 
pute a sufficient number of points on the curve by the elementary 
method (§ 173, p. 266) to give a fair idea of the locus, and sketch 
the curve through the points. 

Ex, 1. Trace the curve y^ = sb*. 

Solution, Let us examine the curve in the above order. 

(a) The curve is symmetrical with respect to OX. 

(b) The curve is not symmetrical with respect to OT, 

(c) The curve is not symmetrical with respect to the origin. 

(d) It passes through the origin. 

(e) Its intercepts on the axes are both zero. 

y > (f) -p = — - ; showing that above OX the curve always has a 

positive slope, and below OX a negative slope. 

It has no maximum or minimum points. 
3 

hence ^bove OX the curve Is concave up- 




wards and below OX concave downwards. There are no points 
of inflection. 

(h) The curve has a cusp of the first kind at the origin, the 
common tangent coinciding with OX, 
(i) There are no asymptotes. 
(i) y = ± V^; hence the curve does not extend to the left of OT, since nega- 
tive values of z make y imaginary. When x = oo, y = db oo, showing that there are 
two infinite branches, one on each side of OX. Plotting a number of points and 
sketching in the curve, we get the semicubical parabola shown in the figure. 

Ex. 2. Trace the curve y' = 2 ox^ — x". 

Solution. This curve is found to be not symmetrical vnth respect to either axis 
or the origin, but it passes through the origin and in addition has the intercept 2 a 



4a. safe 



uii \y.^L.. 




dy 


4 


ox -3x2 

V 








dx 




3y2 ' 




hence when 


x = 


4a 
3 


the 


curve has the 


maximum 


ordinate } a 


^i. 












d^j 




8aa 





^^ 9xi(2a-x)*' 

hence x = 2 a gives a point of inflection on OX, 
to the left of which the curve is concave down- 
wards and to the right concave, upwards. The 
curve has a cusp of the first kin'd at the origin, 




CURVE TRACING 



269 



the common tangent coinciding with OT. The only asymptote is (^£ in figure) 

which lies to the right of the infinite branch in the second quadrant and to the left 

of the infinite branch in the fourth quadrant. From y = V2 ax^ — x* we plot addi- 
tional points and draw the curve shown in the figure. 

175. Tracing of curves given by equations in polar coordinates. 

The rudimentary method is to solve the equation for p when pos- 
sible, assume values for 0, calculate the corresponding values of p, 
plot the points thus found, and draw a curve through them.* 
This work may be facilitated by examining the curve for asymp- 
totes and by noting the values of which make p a maximum or 
a minimum. 



Ex. 1. Trace the curve p = 10 sin 2 $, 



Solution. Tabulating 
p. 4), we have 






Bin2e 


P 


0^ 








16<» 


.60 


6.0 


80° 


.87 


8.7 


460 


1.00 


10.0 


GOP 


.87 


8.7 


lb"" 


.60 


6.0 


90*» 








106° 


- .60 


- 6.0 


120° 


- .87 


- 8.7 


136° 


-1.00 


-10.0 


160° 


- .87 


- 8.7 


166° 


- .60 


- 6.0 


180° 








196° 


.60 


6.0 


210° 


.87 


8.7 


226° 


1.00 


10.0 


240° 


.87 


8.7 


266° 


.60 


6.0 


270° 








286° 


- .60 


- 6.0 


800° 


- .87 


- 8.7 


316° 


-1.00 


-10.0 


330° 


- .87 


- 8.7 


346° 


- .60 


- 6.0 


360° 









the corresponding value of $ and p for every 16* (see table, 

p= 10sin2^. 

p is a maximum when sin 2 ^ is a maximum, and 
this occurs when sin 2^ = 1, or = 46°, 226°, etc. 
This maximum value of p is then 10. 

p is a minimum when sin 2 tf is a minimum, i.e. 
when sin 20 = — 1 or 0= 136°, 816°, etc. Hence the 




minimum value of p is - 10. When = 0, 180°, 
etc., p = 0. If we in addition remember that sin 2 
is a periodic continuous function of 0, it is not neces- 
sary to tabulate many values of and p. The curve 
consists of four loops, as shown in the figure, and 
it is for this reason sometimes called a four-leaved 
rose. 



* The author ha» designed plotting paper for polar oo({rdlnates on which concentric circles 
and radial lines are drawn in faint blue ink. This paper is desirable for the rapid and accurate 
plotting of polar cttrres. Published by the Yale Cooperative Corporation, New Hayen, Conn. 



270 



DIFFERENTIAL CALCULUS 



Trace the following curves. 

1. ya(2a-x) = «». 

2. (x2 + 4aa)y = 8a«. 

3. ai^ = x«-6x«. 

4. (y-x)2 = x«. 
6. x' + y' = a*. 

6. aJV = (6* - y*) (a + y)« 

7. y = log X. 

8. y = «-**. 

10..y=:(xa-l)a. 



11. y = Binx. 

12. y = tan x. 

13. x»(y - o) = a' - xy«. 

14. x*-2ax»y-axy« + aV = 
16. (x2 + y«)» = aa{as«-y«). 

16. p = a cos 2 e. 

17. p = a Bin 3 ^. 

18. p = a (1 — cos &). 

19. p = asin'-. 

3 

20. p = a8ec>-. 

3 



CHAPTER XXII 



APPLICATIONS TO GEOMETRY OF SPACE 



tA«»v-fAy,s-fAx) 



176. Tangent line and normal plane to a skew curve whose equa- 
tions are given in parametric form. The student is already familiar 
with the parametric representation of a plane curve. To extend 
this notion to curves in space, let the coordinates of any point 
(x, y, z) on a skew curve be given as functions of some fourth 
variable t, thus, 

(A) x^4>(t), !/=ir{t), z^xi^y 

The elimination of the parameter t between these equations two 
by two gives the projecting cylinders 
of the curve on the coordinate planes. 

Let the point F (x^ y, z) correspond 
to the value t of the parameter, and 
the point F'{x + Ax, y + Ay, z + As) 
correspond to the value f + A^; t^ 
Ay, As being the increments oix^y^z 
due to the increment Af as found 
from equations (A). From Analytic 
Geometry we know that the direction 
cosines of the secant (diagonal) FF^ are proportional to 

At, Ay, As;; 

or, dividing through by A^ and denoting the direction angles of 

the secant by a\ /8', 7', 

^, , At Ay As 

cos a' : cos /8' : cos 7' : : -— ■ : -77 : -— • 

A^ A^ A^ 

Now let P' approach F along the curve. Then Ae, and therefore 
also Aa:, Ay, A«, will approach zero as a limit, the secant FF^ will 
approach the tangent line to the curve at P as a limiting position, 

and we shall have 

_ dx ay dz 

cos a : cos p : cos 7 : : — : -^ : — » 

271 




272 



DIFFERENTIAL CALCULUS 



where a, /8, 7 are the direction angles of the tangent (or curve) 
at P. Hence the equations of the tangent line to the curve 

at the point (Zj y, z) are given hy 



(67) 



<fag dy dz ' 

dt dt dt 



and the equMion of the normal plane, i.e. the plane passing through 
(a:, y, z) perpendicular to the tangent, is 



m 



g(X-a:) + g(r-y) + g(Z-z)=:0, 



X, Yy Z being the variable coordinates. 

Ex. 1. Find the equations of the tangent and the equation of the normal plane 
to the helix* (6 being the parameter), 

X = acos^, 
^ = a sin ^, 

(a) at any point ; (b) when 9 = 2 r. 



SohUion, 



dz . - dy ^ dz ^ 

= — a sm ^ = — y, -^ = a cos = x. — = 6. 

d$ 



de de 

Substituting in (67) and (68), we get at (x, y^ z), 

X^x T-y Z-z ^ ,„ 

= = — : — » tangent line ; 

— y X b 

and - y (X - x) + x(I^- V) + b{Z - z) = 0, 

normal plane. 

When tf =2 r, the point of contact is (a, 0, 2 bw), 

Pvi"« X-a _ T-0 __ Z-2bK 

" a " b ' 

« 

or, X = a, 6F = aZ — 2 abx, 

the equations of the tangent line ; and 

aF-fftZ -26aT = 0, 
the equation of the normal plane. 



* The helix may be defined as a cunre traced on a right oiroular cylinder so as to cut all the 
elements at the same angle. 

Take QZ as the axis of the cylinder and the point of starting in OX at Pq, Let aaradlus of 
base of cylinder and wangle of rotation. By definition, 

PN PK z , ,, ^ 

■ ""c^r = :r 7rT-= "^ = *' (const.), or , z = ake. 
Sn aro P„y a6 

Let aA-=» & ; then z = b0. Also, y = xMX= a sin 0,x- OM ~ a cos A 




APPLICATIONS TO GEOMETRY OF SPACE 273 

177. Tangent plane to a surface. A straight line is said to be 
tangent to a surface at a point F if it is the limiting position of a 
secant through F and a neighboring point P' on the surface, when 
P' is made to approach P along the surface. We now proceed to 
establish a theorem of fundamental importance. 

Theorem. All tangent lines to a surface at a given point * lie in 
general in a plane called the tangent plane at that point. 

Proof Let 

{A) F{x,y,z)^0 

be the equation of the given surface, and let F{x^ y, z) be the given 
point on the surface. If now P' be made to approach P along a 
curve C lying on the surface and passing through P and P', then 
evidently the secant FF^ approaches the position of a tangent to 
the curve C at P. Now let the equations of the curve C be 

{B) x=.4>{t), y = ir(t), ^ = X(0. 

Then the equation (A) must be satisfied identically by these 
values, and since the total differential of (A) when Xy y, z are 
defined by {B) must vanish, we have 

(C) 7-H f--\ r = 0- (50), p. 199 

^ ^ dx dt dy dt dz dt ^ ^' ^ 

This equation shows that the tangent line to C, whose direction 
cosines are proportional to 

dx dy dz 

It' w w p- ^^^ 

is perpendicular f to a line whose direction cosines are determined 

by the ratios 

dF dF dF 

^_- • —— * » 

dx' dy' dz 
and since C is any curve on the surface through P, it follows at 

• The point in question is assumed to be an ordinary (non-singular) point of the surface, i.e. 

fijfi ^w ^f 

-—«—-»—- are not all zero at the point. 
ox cy dz 

t From Solid Analytic Geometry ve know that if two lines haying the direction cosines 
oosai, cot^xi <i08Yi and cos a,, cos^t, cosy, are perpendicular, then 

cos Ox cos oj + cos Px cos /3t + cos 71 cos 7,= 0. 



274 DIFFERENTIAL CALCULUS 

once, if we replace the point P {x^ y, z) by F^ (a^j, y^, z^^ that all 
tangent lines to the surface at P^ lie in the plane* 

which is then the forrmdafor finding the equation of a plane tangent 
at (oTj, y^, z^ to a surface whose equation is given in the form 

F{x,y,z)^0. 

In case the equation of the surface is given in the form z =/(a:, y), let 
(2>) P(a:,y, z)=/(2:, y)-z = 0. 

Th n dF^d£^dz_ dF^^^dz ^^_^ 

dx dx dx cy dy dy dz 

If we evaluate these at {xy, y^, z^ and substitute in (69), we get 

which is then the formula for finding the equation of a plane tangent 
at (Xj, y^ 2j) io a surface whose equation is given in the form z=if(Xj y). 

In § 137, p. 200, we found (53), the total differential of a function u (or z) of x 
and V, namely, 

(E) dz = — cte H — dy. 

^ ' dz ty 

We have now a means of interpreting this result geometrically. For the tangent 
plane to the surface z =/(x, y) at (x, y, z) is, from (70), 

(F) Z-2 = ^(X-x) + J?(r-y), 

oz cy 

Xy F, Z denoting the variable co-ordinates of any point on the plane. If we 

substitute jr = * + dxandr=» + (ly 

in (F), there results 

(G) Z'-z = ^dz + ^dy. 

^ ' dz dy ^ 

Comparing {JE) and (G^, we get 

(H) dx^Z-z, Hence 

• The direction coeines of the normal to the plane (69) are proportional to ^*, ^i ^. 

dX) oyx dzi 
Hence from Analytic Geometry we see that (C) is the condition that the tangents whose direc- 
tion cosines are cos a, cos^, cos y are perpendicular to the normal ; i.e. the tangents must lie in 
the plane. 

t In agreement with our former practice, 

dFi dF^ dF^ dz^ dz^ 
d^x dvi C2, dxj dpx 
denote the values of the partial derivatiTes at the point (x^, y^, s,). 



APPLICATIONS TO GEOMETRY OF SPACE 275 

The total differential of a function f{x, t/) corretponding to the incre- 
menit dz and dp equtdi the eorreipondin^ inerement (if the t eoordtnate (>f the (anient 
plane to the tur/ace t =f(z, j/). 



Thus, In the figure, PP" U the pl&ne tangent t« nuiace PQ &t P(x, y, i). 
Let AB = dx and CD = dj/; 

then di = Z-t = DP"- DE = EP". 

178. Nonnal line to a sorface. The normal line to a surface at 
a given point is the line passing through the point perpendicular 
to the tangent plane to the surface at that point. 

The direction cosines of any line perpendicular to the tangent 
plane (A9) are proportional to 

e/\ £^ e^ 

^ ' dF\ BFx BF\ 

dxi dffx &*i. 

are the equationt of the normal line • to the tnrface F{x, y, z) = at 

Similarly, from (J#), 

t>xi dyi 

are the equatwnt of the normal line* to the surface Zi=f{x, y) at 

• See lootnole. p. ZT4. 



276 DIFFERENTIAL CALCULUS 



EXAMPLES 

1. Find the equation of the tangent plane and the equations of the normal line 
to the sphere x> + 2^ + 2^ = 14 at the point (1, 2, 3). 

Solution. Let F{x, y, «) = x» + y^ + «« - 14 ; 

dF dF dF 

then — = 2x, — = 2 y, — = 2 z ; Xi = 1, yi = 2, «i = 3. 

dz dy dz 

dxi dyi dzi 

Substituting in (60), 2(x- 1) + 4(y- 2) + 6(z -3) = 0, x + 2y + 3z = 14, the 
tangent plane. 

X— 1 y-2 z-3 



SubsUtuting in (71), 



'» 



2 4 6 

giving z = 3 X and 2 z = 3 y, equations of the normal line. 

2. Find thd equation of the tangent plane and the equations of the normal line to 
the ellipsoid 4x> + 9y9 -f- 36 z^ = 36, at point of contact where x = 2, y = 1, and z 
is positive. ^n«. Tangent plane, 8(x - 2) + 9(y - 1) + 6 Vll(z - J Vil) = 0, 

... x-2 y-1 z-jVu 

normal hne, = ~ = =-==— 

* 8 9 6vTl 

3. Find the equation of the tangent plane to the elliptic paraboloid z = 2 x> + 4 y* 
at the point (2, 1, 12). Am. 8x + 8y-z = 12. 

4. Find the equations of the normal line to the hyperboloid of one sheet 
«" - 4y» + 2z2 = 6 at (2, 2, 3). Ana. y + 4x = 10, 3x - z = 3. 

5. Find the equation of the tangent plane to the hyperboloid of two sheets 

6. Find the equation of the tangent plane at the point (Xi, yi, zi) on the surface 
ckb' + 6y* -f- cz« + d = 0. Ans. axiX + byiy + cziZ + d = 0. 

7. Show that the equation of the plane tangent to the sphere 

x* + y' + 2* + 2Lx-f2 3fy + 2^z + D = 
at the point (xi, yi, Zi) is 

xix + yiy + 2i2 + i(x + Xi) + M(y + vi) + N(z + 21) + i) = 0. 

8. Find the equation of the tangent plane at any point of the surface 

«' + y' + 2' = a', 

and show that the sum of the squares of the intercepts on the axes made by the 
tangent plane is constant. 

9. Prove that the tetrahedron formed by the co5rdinate planes and any tangent 
plane to the surface xyz = a^ is of constant volume. 



APPLICATIONS TO GEOMETRY OF SPACE 



277 



179. Another form of the equations of the tangent line to a skew 
curve. If the curve in question be the curve of intersection AB 
of the two surfaces F(xy y, 2) = 
and G (2;, y, z) == 0, the tangent line 
FT at JP(ari, y^, z^) is the intersection 
of the tangent planes CD and CF at 
that point, for it is also tangent to 
both surfaces and hence must lie in 
both tangent planes. The equations 
of the two tangent planes at F are, 
from («9), 




fdFx, V , dFx, . , 

(a? — Xx) + -r—(y - Vi) + 



(78) 



dxx 
dOx 



dyx 



dFx 
dzx 



(z - zx) = O, 



(05 - 051) + ^\y -yi) + ^(Z- Zx) = O. 



dxx ^ '' ' dyx ^^ "" ' dzx 

Taken simultaneously, the equations (78) are the eqiuitians of 
the tangent line FT to the skew curve AB, Equations (78) in more 
compact form are 

35 — 051 y — yi « — Zx 



(M) 



dFxdGx dFxdGx dFxdGx dFxdGx dFxdOx dFxdOx 
dyx dzx dzx dyx dzx ctei dxx dzx dxx dyx dyx dxx 



or, using determinants, 

05 — 051 



(W 



_ y — yi _ as — gj 



dFx dFx 
dyx dzx 

dGxdOx 
dyx dzx 




dFx dFx 
dzx dxx 

dGxdOx 

dzx dxx 




dFx dFx 
dxx dyx 

dOxdGx 
dxx dyx 



180. Another form of the equation of the normal plane to a skew 
curve. The normal plane to a skew curve at a given point has 
already been defined as the plane passing through that point per- 
pendicular to the tangent line to the curve at that point. Thus, 
in the above figure, PHI is the normal plane to the curve AB at P. 
Since this plane is perpendicular to (75), we have at once. 



(W 



dFx dFx 



dyx dzx 
dGxdOx 



dyx dzx 



(X — 05i) + 



dFx dFx 



dZx dxx 
dGx dGx 



dzx dxx 



(y - vi) + 



dFx dFx 



dxx dyx 
dGx dGx 



dxx dyx 



(z - Zx) = O, 



the equation of the normal plane to a skew curve. 



DIFFERENTIAL CALCULUS 



BZAKPLE3 



1. Find the eqafttiooa of the tangent Une and the equation o( the normal plane 
at (r, r, r W) to the curve of intersection of the sphere and cjlinder whose equa- 
tions are reapecUvel; ^ + j/* + z' = 4t', x* + v' = 2tx. 



Solution. LetF=i' + v* + «'-4r' and O^xf + y'-in. 

dXi dj/i dXi 

?°! = 0, ?°! = 2r, ?2! = 0. 

SX) tgi dti 

Substituting in (76), ^-^ = '^^^ = ^^^ ; 

— ^2 1 

or, p = r,a + v'2j = ar, 

the equations of the tangent PT at P to the curve of intersection. 
Substituting in (76), we get the equation of the normal plane, 
- v^ (I - r) + (1/ - r) + (I - r Ve) = 0, 
or, V2x-z = 0. 

2. Find the equations of the tangent line to the circle 

»' + y* + *" = 26, 
a + I = 5, 
at the point (2, 2 Vs, 3). ^^ 2^ + 2 Vl:, + 8r = 26, . + . = & 

3. Find the equation of normal plane to the curve 

ji + ^ + iS = ra, 

iS _ n + yi = 0, 

"tfJi, 1/1, Zi). Ana. 2yiz,x - {2xi ~ r)z,v - rtfit = 0. 



APPLICATIONS TO GEOMETRY OF SPACE 279 

4. The equations of a helix (spiral) are 

«2 + y« = r«, 

y = xtan-* 
c 

Show that at the point (asi, yu Zi) the equations of the tangent line are 

c(x-xi)-f-yi(z-«i) = 0, 
c(y-yi)-xi(«-2i) = 0; 

and the equation of the normal plane is 

ViX — Xiv — c (z — «i) =s 0. 

x' t/* z^ 

5. A skew curve is formed by the intersection of the cone — + tt — :; = 

' a^ t^ c^ 

and the sphere x^ + ^ + a^ = r^. Show that at the point (xi, yi, Zi) the equations 
of the tangent line to the curve are 

c^{a^ - 6*)xi(x ^ xi) = - a»(6a + c«)zi(z - Zi), 
c'^ia^ - 62) yi (y - yi) = + 6»(c« + aa)zi (z - Zi) ; 

and the equation of the normal plane is 

a»(6» + c») yiZiX - 6^ (c« + a^) ziXiy - c^ (a* - 6^ XiViZ = 0. 



CHAPTER XXin 



CURVES FOR R£F£R£irC£ 



For the conyenience of the student a number of the more 
common curves employed in the text are collected here. 



Cubical Parabola 
Y 



Semicubical Parabola 





y = ax*. 



t^ = ax\ 



The Witch of Aqnesi 



r 


1 




-^K 


'J 


^"^^ 


o 




X 



xhj=z4^a^(2a^y). 



The Cissoid of Diocles 




y^(2a — x) — x\' 
p = 2 a sin tan 9. 



280 



CURVES FOR REFERENCE 



281 



The Lemniscate of Bebnoulli The Conchoid of Nigomedes. 




p* = a* cos 2 $. 





p = a sec $:tb. 



Cycloid, Ordinary Case 




Cycloid, Vertex at Origin 

Y 




x=:a arc vers 



{:: 



^-V2ay-y». 
a 



a (^ — sin 6), 
a(l — cos^. 



X = 



t: 



a arc vers - -h '>J2ay 
a 

a (^ + sin ^), 
a(l — cos fl). 



-y^ 




Parabola 



y=5(^ + ^"")- 




as* + y* = a*. 



1 



282 



DIFFERENTIAL CALCULUS 



Hypocycloid of Fouk Cusps 



EVOLUTB OF ElLIPSB 




05* -f y* = a*. 
' x=za cos'^, 




(ooj)* + (hy)^ = (a" - h^K 



Cakdioid 



Folium of Descabtes 





35' -f y' + «a5 = a Vx* -f y\ 
p = a (1 — cos ^). 



«' -h y* — 3 oay = 0. 



CURVES FOR REFERENCE 



283 



L1MA90N 





p=zh — a cos 0. 



yl J— X* 



Spiral of Asghimedss 




Logarithmic or Equiangular 

Spiral 




= aft 



p = «*»•, or 
log p = a0. 



Hyperbolic or Reciprocal 
Spiral 



L1TUU8 




pO=ia. 




pV = a*. 



284 



DIFFERENTIAL CALCULUS 



Fababolig Spiral 



Logarithmic Curve 




^, -" 




(p--a)" = 4a(j^. 



y = log X. 



Exponential Curve 



Probability Curve 





y = «'. 



y = e"*^. 



Sine Curve 



Tangent Curve 







o 



y = 8in X, 



y = tana;. 



CURVES FOR REFERENCE 



285 



Thbee-Leaved Rose 



Three-Leaved Rose 





p =za sin 3 $, 



p = a cos 3 $, 



Four-Leaved Rose 



Four-Leaved Rose 





p = a sin 2 $. 



p = a cos 2 B. 



INTEGRAL CALCULUS 



CHAPTER XXIV 

INTEGRATIOir. RULES FOR DrTEGRATDrO STANDARD 

ELEMENTART FORMS 

18L Integration. The student is already familiar with the 
mutuallj inverse operations of addition and subtraction, multipli- 
cation and division, involution and evolution. In the examples 
which follow the second members of one column are respectively 
the inverse of the second members of the other column. 

y = a? + l, a; = ± Vy-1; 

y zzzamccj x=z arc sin y. 

From the Differential Calculus we have learned how to calcu- 
late the derivative f{x) of a given function /(x), an operation 
indicated by ^ 

or, if we are using differentials, by 

df(x)=:f{x)dx. 

The problems of the Integral Calculus depend on the mver%e 
operation^ namely : 

To find a function f{x) whose derivative 

(A) f(x) = 4>{x) 
is given. 

Or, since it is customary to use differentials in the Integral 
Calculus, we may write 

(B) df{x):=.f(x)dx=^<l>{x)dxy 

and state the problem as follows : 

Saving given the differential of a function^ find the function itself, 

287 



288 INTEGRAL CALCULUS 

The function /(a;) thus found is called an integral* of the given 
differential expression, the process of finding it is called integratwn^ 

and the operation is indicated by writing the integral sign t j in 
front of the given differential expression. Thus, 

(O) ff{x)dxt^f{x), 

read, an integral of f{x)dx equals f{x). The differential dx indi- 
cates that X is the variable of integration. For example, 

(a) If f{x) = a?y then f(x)dx=^^ a^cfa, and 

(b) If f{x) = sin x, then f (x) dx = cos xdx^ and 



/ 



cos xdx = sin x. 



dx 
(c) If /(a:) = arc tan a:, then f(x)dx==- — -^» and 

dx 



A 



= arc tan x. 



+7? 

Let us now emphasize what is apparent from the preceding 
explanations, namely, that 

Differentiation and integration are inverse operations. 
Differentiating (O) gives 

(2>) dCf (x) dx =f (x) dx. 

Substituting the value oif{x) dx[=z (if{x)] from (B) in ((7), we get 

(E) jdf{x)^f{x). 

d C 

Therefore, considered as symbols of operation, — and I • . . <2a; 

are inverse to each other; or, if we are using differentials, d and I 
are inverse to each other. 

* Called antv-d\fftreniiaX by some writers. 

t Historically this sign is a distorted <9, the initial letter of the word turn. Instead of defining 
integration as the inrerse of differentiation we may define it as a process of summation, a rery 
Important notion which we will consider In Cliapter XXX. 

X Some authors write this U^ V(^) when they wish to emphasize the fact that it is an inverse 
operation. 



INTEGRATION 289 

When d is followed by j they annul each other, as in (D), but 

when j is followed by (2, as in (E)^ that will not in general be the 

case' unless we ignore the comtanb of integration. The reason for 
this will appear at once from the definition of the constant of 
integration given in the next section. 

182. Constant of Integration. Indefinite integral. From the pre- 
ceding section it follows that 

since d(Qi?) =3 a^dx^ we have | 8 a^dx = a? ; 
since d(2? -|- 2) = 8 a^do^ we have j 3 3?dx = a?" -f- 2 ; 

since d(aj" — 7) = 8 st^dx^ we have j 3 a^dx = a? — 1. 

In fact, since 

d(a:»+(7)=3a?(ir, 

where C is any arbitrary constant, we have 

r3a?(ic = 2^-f C. 

A constant C arising in this way is called a constant of integration* 
Since we can give C as many values as we please, it follows that 
if a given differential expression has one integral, it has infinitely 
many differing only by constants. Hence 



/^ 



f(x)dx=f{x)+C; 

and since C is unknown and indefinite, the expression 

f{x) + c 

is called the indefinite integral off{x) dx. 

It is evident that if ^ (x) is a function the derivative of which is 
f(x)y then ^ (x) -f C, where C is any constant whatever, is likewise 
a function the derivative of which isf{x). Hence the 

Theorem. If two functions differ by a constant, they have the 
same derivative. 

• ConstMit here means that it is Independent of the variahle of integration. 



290 INTEGRAL CALCULUS 

It is, however, not obvious that if ^ (2:) is a function the deriva- 
tive of which is/(2;), then all functions having the same derivative 
f(x) are of the form 

where C is any constant. In other words, there remains to be 
proven the 

Converse Theorem. If two functions have the same derivative^ 
their difference is a constant 

Proof Let ^(x) and -^ (x) be two functions having the common 
derivative f(x) . Place 

F{x) = ^(x)— ylr(x); then 
(A) F' (x) =-§-[<l>{x)^it (x)] =f{x) -/(x) = 0. By hypothesis 

But from the Theorem of Mean Value, (44), p. 168, we have 

F{x^Ax)^F{x)==AxF'(x + 0'Ax). 0<tf<l 

.-. F{x + Ax)-F{x) = 0, 

[Since by (A) the derivative of F(x) ]a zero for all values of x.] 

and F(x + Ax) = F(x). 

This means that the function 

F{x)=.<l>(x)^ir(x) 

does not change in value at all when x takes on the increment Ax^ 
i.e. <f>{x) and '^{x) differ only by a constant. 

In any given case the value of C can be found when we know 
the value of the integral for some value of the variable, and this 
will be illustrated by numerous examples in the next chapter. 
For the present we shall content ourselves with first learning how 
to find the indefinite integrals of given differential expressions. 
In what follows we shall assume that every continuous functi(m 
has an indefinite integral^ a statement the rigorous proof of which 
is beyond the scope of this book. For all elementary functions, 
however, the truth of the statement will appear in the chapters 
which follow. 

In all cases of indefinite integration the test to be applied in 
verifying the results is that the differential of the integral must he 
equal to the given differential expression. 



INTEGRATION 291 

183. Rules for integrating standard elementary forms. The 

Differential Calculus furnished us with a General Rule for differ- 
entiation (p. 42). The Integral Calculus gives us no corresponding 
general rule that can be readily applied in practice for performing 
the inverse operation of integration.* Each case requires special 
treatment and we arrive at the integral of a given differential 
expression through our previous knowledge of the known results 
of differentiation. That is, we must be able to answer the question, 

What function^ when differentiated^ mil yield the given differential 
expression f 

Integration then is essentially a tentative process, and to 
expedite the work, tables of known integrals are formed called 
standard forms. To effect any integration we compare the given 
differential expression with these forms, and if it is found to be 
identical with one of them, the integral is known. If it is not iden- 
tical with one of them, we strive to reduce it to one of the standard 
forms by various methods, many of which employ artifices which 
can be suggested by practice only. Accordingly a large portion 
of our treatise on the Integral Calculus will be devoted to the 
explanation of methods for integrating those functions which fre- 
quently appear in the process of solving practical problems. 

From any result of differentiation may always be derived a 
formula for integration. 

The following two rules are useful in reducing differential 
expressions to standard forms. 

(a) The integral of any algebraic sum of differential expressions 
equals the same algebraic sum of the integrals of these expressions 
taken separately. 
, Proof Differentiating the expression 

\ du-\' \ dv — J dw^ 

w, v, w being functions of a single variable, we get 

du-^dv — dw. III, p. 144 

[1] .-. i(du -f dv — dw) = t du + i dv ^ j dw. 

* Eren though the Integral of a glren differential expression may be known to exist, yet It 
may not be possible for us actually to find it in terms of known functions, because there are 
functions other than the elementary functions whose deriyatives are elementary functions. 



292 INTEGRAL CALCULUS 

(6) A constant factor may be written either btfore or after the 
integral sign. 

Proof Differentiating the expression 

a i dv 

gives adv, IV, p. 144 

[2] •*• I adv = a I dv. 

On account of their importance we shall write the above two 
rules as formulas at the head of the following list of 

STANDARD ELfiMENTART FORMS 
[1] . j (du + cfv — dw) = I du + Idv — j dw. 



[9] i adv = a I dv, 

[8] Cdx = a? + C. 



[4] Cv**dv = ^^^ + C. n=^-l 



[« 



[10] 



n+1 



[5] J^:=logV+C 



= logr V + log c = logr cr. 

[Placing C^logc] 



[6] Ca^dv = -^5l_ + c. 

J log a 

[»] fe^'cli; = c« + C. 

[8] r sin vdv = '- cos v + C- 

I cos vdt; = sin v + C 



I sec* vdr = tan v + C. 
[11] I cosec* v ffv = — cot V + €• 

[19] r secf tanf dt; = secf + C. 



INTEGRATION 293 



[18] 



j C8C V cot t» «f V = — CSC V + C* 

[14] Ctaxkvdv = log sect; + C. 

[15] I cot vdv = log sin v + C 

[16] J sec vdv = logrtan (h + t) + ^• 

[IT] fcsc vdv = log taH (^) + C. 

J t;» + a* a a 



[18] 



[19] 



[20] f ,^^ = arc sin ^ + C. 



[21] 



C—M= = log (V + Vv* ± a«) + e. 



, = arc vers- + C. 

V2av - V* « 

— ^ = — arc sec— + C 



[28] 



Proof 0/ [8]. Since 

rf(a: + C)=<fo, II, p. 144 

we get j dx=:x + C. 

Proof of [4]. Since 

d (J—- + C^ = ifdv, VII, p. 144 

\«-hl / 

J^^n + l 
trdv = 1"*"^* 

This holds true for all values of n except w = — 1. For, when 
» = — 1, [4] gives 

v''dv = Ai T + C = - + C=oo + C7, 

— 1 + 1 

which has no meaning. 

The case when n = — 1 comes under [5]. 



294 INTEGRAL CALCULUS 

Proof of [5]. Since 

d(logt; + C) = -^ IX, p. 145 

we get r — = log v + C. 

The results we get from [5] may be put in more compact form 
if we denote the constant of integration by log c. Thus, 

I — = log V + log c = log cv. 

Formula [5] states that if the eapression under the integral ngn i$ 
a fraction whose numerator is the differential of the denominator, then 
the integral is the natural logarithm of the denominator. 



BXAMPLBS 

For formulas [l]-[6]. 

Verify the following integrations. 

sfidx = — H C, by [4], where v = x and n = 6. 

2. faz^dx = aCz^dx by [2] 

= — + 0. By [4] 

3. J{2s^-6x*---Sx-^4t)dx=j22*dz-j6x^»dt-'Csxdx-^C4dz by [1] 

= 2 fx^dr - bCx^dx - sCxdx + iCdx by [2] 

= 2-X--i- + '^-^^- 

Note. Although each separate Integration reqalres an arbitrary constant, we write down 
only a single constant denoting their algebraic sum. 

4. J(-^-^ + 3cVS2^dx=J'2ax-*dx- ffcx-'dxH- r8cx«dx by [1] 

= 2 ajx-^dx - bCx-^dx + 8 cCx^dx by [2] 

= 2a.?^-5.^ + ?^ + C l^[4] 

= 4aVx + -+fcx* + C. 

X 

6. j2ax*-i(ix = ^ + C. 6. Js m^^dz = ?^ + C. 



INTEGRATION 295 






JBinl. First eziMmd. 

(nx) • 

13. /(a. + M«V«ix = <^±^ + C. 

Hint, ThiB may be brought to form [4]. For let v^a'+Mz* and «»); then <iv»2M«df. 
If we now insert the constant factor 26* before xdx, and its reciprocal — before the integral 

sign (so as not to change the ralue of the expression), the expression may be integrated, using 
[4], namely, 

v^dv=- + C. 

n + 1 

Thns, r(a« + 6»jB«)*«ite«-^ r(a»+6«a:^*26«a:d;t--^ r(a» + 6»JB«)*<l(a> + M«^ 

"26«' I '^^" 36« "*'^' 

JVbf«. The student is warned against transferring any function of the Tarlable from one tide 
of the integral sign to the other, since that would change the value of the integral. 

14. fVa* - x^xdz = - J(a2 - x«)i + C. 

15. C{Sax^ + 4te«)*(2(ix + 46a;«)dx = }(8axa + 4te«)* + C. 
Hint. Use [4], making va3ax*+46a;* and nB|. 

16. r&(6axa + 86x»)«(2ax + 4ea«)dr = — (6ax« + 865e«)* + C. 
•/ 16 



!''• f . .i = i(«' + g')* + <^- 
J (a« + x«)* 

J7i»<. Write this C(a^ + a*)" * afldx and apply [4]. 

18. f ^^. = -2Vl-a-|.C. 
^ Vl-x 

19. j2ry(^ + l)*dy = |^(ya + p«)t + C. 



296 INTEGRAL CALCULUS 



20 f ^a^da ^ a 



Solution. 



Saxdx ^ r xdx 



ri^axax ^ r xax 

J 62 + c2x« J 62 + e^« ^^^ 

This resembles [5]. IVir let o«6*+e^*; then dv=2^xdx. If we introdnoe the faetor 2e* 
after the integral sign, and — - before it, we haye not changed the yalue of the expression, but 
the numerator is now seen to be the differential of the denominator. Therefore 

r Xdx ^8a/-2c^^3a/-d(62 + e^')^8a 
J 62 + 62x2 2d2j 62 + ^2 2e2J 62 + e2x2 2e2 ^^ ^ '^ ^^^ 

22. r-|^ = llog(x2-l) + logC = logC Vx23i. 
•/ x* — 1 • 

-_ r 56xdx , c 

J8a-66x2 (8a-66x2)* 

26. r-^ = x-^ + ^-log(x + l) + a. 
•/ X 4* 1 2 o 

^«<. First divide the numerator by the denominator. 

27. r?^^dx = x -log(2x + 8)2 + C. 
J 2x + 3 

28. r^!lillldx = ilog(x«-nx) + C. 

/fti-icK 1 
a + 6c* ?io 

31. J (log a)8 ^ = J (log a)* + C. 



32. 



t2 + 1 



. f^— ti(fr = ^ + r + 21og(r-l) + a. 
J r — 1 2 

J (a + 6M»)"» 6n(l — m) 

Proofs of [6] and [1]. These follow at once from the corre- 
sponding formula for differentiation, X and Xa, p. 145. 



INTEGRATION 297 



For formulas [6] and [7]. 
Verify the following integrations. 

J 2 log a 

8oi.Miicm, Cb(fi*dx = bfa^'dx. By [2] 

Tbla reeembles [6]. Let r b2x ; then dv'^2dx. If we then insert the factor 2 before dx and 
Jie factor \ before the integral sign, we hare 

6/«..dx = ^/a.^<fc = 5ja.^d(2«) = ^.j^+C. B, [6] 

4. J c"dte = ne» + C 

6. r6^+**+«(x + 2) dx = ief^ + te + « + C. 

6. ((a^-lF^)dx = -^ — r + C. 

J n log a m logo 

7. (a'tFdx^—^^ + C. 

J 1 + log a 

(e« 4- e •)dx = a(€« - e «) + C. 
11. J.^ = ^ + c- + log(c--l) + C. 



12. 



•e^-1 



r?— idr = log(e^ + 1)2 - r + C. 
J er + l 



13. r ('"-f)' dx= °'^'^,°-t -2x+c. 

J a*&^ log a — log 6 



Proofs of [8]-[18]. These follow at once from the corresponding 
formulas for differentiation, XII, etc., p. 145. 



298. INTEGRAL CALCULUS 



sin vdv /•— sin vdv 



Proof of [14]. ftan vdv = C?^^ = _ f 

•^ •/ COS V J 



COS V 

/ d (cos ty) 
COS V 

= -logcosv + C by [5] 

= log sec V -f- C 

[3i]ioe-logQOflVi--log "-lQgl + logaeoi7»loffieov.] 

Pro./ of [15]. foot ri« = f^^^ = f-^i^illi^ 

•^ •/ Sin V J sin v 

= log sin t; + C. By [5] 



Proof of [17]. Since esc t; = esc v 



CSC V — cot V 
CSC V — cot V 



•^ •/ CSC V — cot v 

_ (* d{(iSG y — coty) 
CSC V — cot V 



r 

= log (CSC v — cot v) -f. C by [5] 

, /I — cos v\ . ^ 

= log( — : ] + C 

\ sm t; y 

2sin»^ 

=iog — ; — v'^^ ^y ^'^' P- ^' ^^' p- ^ 

2 sin- cos - 
= logtan^ + C. 

Pr<?<?/ 0/ [16], Substituting v + ^ for v in [17] gives 

J'csc U + 1^ rft> = log tan ^1 + 1 V C. 
But cscf t; + ^ j = seow; therefore 

J aeo vdv = log tan /^| + 1'\ + c. 



INTEGRATION 299 



For fornralas [8]-[17]. 

Verify the following integratiomu 

1 r • o J coB2ax , ^ 

1. I sm 2 axdx = h C, 

J , 2a 

Solution, This resembles [8]. For let v « 2 ox ; then dv « 2 ckte. If we now insert the factor 
2a before dx and the factor — before the integral sign, we get 

/Bin 2 axdx = — I sin 2 ax* 2 adi 
2aJ 

\ r 1 

= — I Bin 2 ox -(2(2 ox) = cob 2 ox + C by [8] 

2aJ ^ 2a *■ '' 

coB2aa; ^ 
2a 

2. fcoBmxdserr— Binnix + C 3. IdBec'tedx = -tan&G6 + C 
J m Jo 



4. r^coB--8in8tfW = 38in- + }co88tf + a. 

5. r7 8ec8atsn8ada = }8ec8a + C. 

ik cos (a + by)dy = ^sin (a + ^) + C 

7. rcoaec«a5»-««dx = - Jcotx« + C. 

8. r 4 C8C ox cot axdx = — esc ox + C^> 
•/ a 

. r Binxdx , c 

9. I — = log 

J a + & COB X 



1 
(a + & cos x)h 







. re«»*Binxdx = -c«»* + C. 

^' f «/^ _ =-^tan(a-6x) + C. 
J COS* (a — te) 

2. fcosnogx)— = 8inlogx + C. 14. f <L±^?^5l^ = log (x + gin x) c. 
•/ . X •/ X -f Bin X 

3. r-=^ = -ncot- + C. 16. I — rrr=:8ec0 + C. 
J ^j^,x n J co8>0 

n 
6. r (tan a + cot a)^da = tan a - cot o + C. 



800 INTEGRAL CALCULUS 

17. r(86C/9-tan/3)3d/3 = 2(tan/9-8ec/3)-/3 + C. 

ai+iintf COS ede = 4- C. 

log a 

19. r(tan2u-l)*du = ^tan2u + logco82tt + C. 

«^ r. . . . vj log (1 — tan a tan y) , ^ 

20. I tan y tan (y + a)cl2^ = - y ^^ ^ + C. 

J tana 

Proof of [18] . Since 
.. i ("- aro tan - + (7^ = - ^7 =-^, by XIV, p. 146 



© 



rfv 1 . V 



we get r -r r = - arc tan - + C* 

Proof 0/ [19]. Since -5^, = ^ f-^^ —\ , 

cj^^:lc(j, ?_v. 

•/ v^ — c? 2aJ yv — a v-\-aJ 

= 2^11og(v-«)-log(t; + a)KC by [5] 

2a v-f-a 
Pro^ of [20]. Since 

<-) 

rf/^arosin-H-cV . ^''^ ^ rft. ^ by XIX, p. 145 
we ffet I — , = arc Bin - + (;• 






•Atoodl-arccot-+ C'|= — — — ; and I --—,= - -arc cot- + C Henoe 



- arc tan - + C=» — arc cot - + C. 



v* + a* a a a a 

Since arc tan - + arc cot - = - 1 we see that one result niay be easily transformed into the other. 

a a 2 \ ^ 

The same Icind of discussion may be given for [20] inyolying arc sin - and arc cos -i and for [23] 
InTolTing arc sec - and arc oc - . 



INTEGRATION 801 

Proof of [21]. Assume Vt;*-f.a* = 2, a new variable. 
Then ^ -{-€?=. ^^ and differentiating, 2 vdv z=. 2 2k2a;, or, 

dv dz 

Z V 

By composition in proportion, 

dv dz dv 4- dz 



therefore 



Z V V -{-z 
dv dv + dz 



Vt^Tf^ v^-z 



[EeplaolBg s by Its equal V»«+o«.] 

Hence ^ r^^^^ /-rf^:^ 

= log(t; + g) +^ by [5] 

= log(v + Vv* + a*) + C. 

In the same way by assuming 

Vv* — a' = 2, 

we get f . / , = log(v + y/v'-a^ + C. 

•^ V v'* — a 

Proofs of [%i] and [38]. These follow at once from the corre- 
sponding^ formulas for differentiation, IXIII and IIY, p. 48. 

EXAMPLES 

For formulas [18]-[23]. 

Verify the following integrations. 



^ r dx 1 2x ^ 

1. I — - — - = - arc tan -— + C. 
-^ 4x«4-9 6 8 



SolutUm, This resembles [18]. For, let v^z=4x^ and a< = 0; then v=:2x, 
do = 2 dx, and a = 8. Hence if we multiply the numerator by 2 and divide in 
front of the integral sign by 2, we get 



/ dx __ 1 r 2dx _ 1 r d{2x ) 
4x2 4-9 " 2 J (2x)« -f (3)2 " 2 J (2x)2 + 

1 2x 



(3) 



2 



=r - arc tan —- + C. By [18] 

6 8 






802 INTEGRAL CALCULUS 

6. f_jL= = hog(6y+V^3;^ + a. 

6. I _ = - arc am z« + C 8. I — , = - arc aec •— + C 

•^ Vl31? 2 -^ xV4x«-9 3 8 

7 f — — - arc tan — 4- C 9. I . = arc vers - + C. 

^ r edt e, W + a.^ 

/• 7d» 7 . ^/6 , ^ 

1. I , z = —^Mcsm\- 8-^-0, 

, /• coBcKia 1 . /8ina\ , ^ 
^- I -s TT- = - arc tani ) + C. 

I. r = arc sin c* + C 

>. I — ^ = arc sin (log x) + C 

•^ « Vl - logsx 

6. I ^ = arc sin \- C. 

•^ Vaa - (u + 6)« « 

_ f adz a ^ 2- c , _ 

7. I — r- = 7 arc tan — h C, 

^ r dx 1 x + l^ 

8- |-^ — ^ ^= arctan-^ + C. 

J x2 + 2x + 5 2 2 

Ainf. By oompletiog the square In the denominator this expression may be brought to a 
form similar to that of Ex. 17. Thus, 

-si =z \ »-arctan + C. By [181 

a4 + 2j: + 6 J (x« + 2x + l)+4 J(x+l)« + 4 2 2 '^ ^ 

Here 9 » a; + 1 and a* 2. 

/nn^ Bring this to the form of Ex. 16 by completing the square. I!hu8, 

/dx r dx r dx C dx . 2x-l ^ _ ^^, 

, "1 . =) , ^ ==\- -arcsin— -— +C. By [20] 



X9. I — = arc sin — \- C. 



Here v = x-^ and a« f . 



20. f ^ =-4arctan^^ + C, 
J 1 + X + x^ 



INTEGRATION 803 



dz 



21. r = = arcBin(2g-3) + (7. 

23. f ^y =2-iog^y + «-^^a 

26. fr-r — T - = arctan(2«-l) + a 

26. f ^ : = log(< + a-f V2cM-H«)-f C> 

•^ V2a« + ^ 

€Wf C ^ I CX . ^ 

27. I — 7=== = — axx5 8ec-7 + C. 
•^ X Vc^c* - a^fra act ao 

28. r-4^= = iarcyerBl8x« + C. 
•^ Vx«-9x« 3 

•/ a^ + X" a a 2 

30. r^^"^dx = llog(3x»-2)--^log^"^-^ + C. 

184. Trigonometric differentials. We shall now consider some 
trigonometric differentials of frequent occurrence which may be 
readily integrated by being transformed into standard forms by 
means of simple trigonometric reductions. 

Example I. To find I sin"* a; no^xdx. 

When either m or n is a positive odd integer, no matter what 
the other may be, this integration may be performed by means 

of formula [4], ^+i 

I ifdv = -' 

J n + l 

For the integral is reducible to the form 

I {terms involving only co%x)9inxdx^ 
when sin x has the odd exponent, and to the form 

I (terms involving only sin x) cos xdx^ 



304 INTEGRAL CALCULUS 

when cos x has the odd exponent. We shall illustrate this by 
means of examples. 

Ex. 1. Find rsin^xcoB^xdas. 

iSoZution. fsin^x cos^ xdx = Isin^x cos^ x cob x(2x 

= fsin^x (1 - 8inSx)s cos xdx 28, p. 2 

= r (sin^x — 2 Bin^x + sin^x) cos xdx 

= Msin x)> COB xdx — 2 Main x)^ cos x^x + j (sin ^^ cob xdx 

sin'x 2 8in^x Bin^x ^ « -„ 

, o o 7 

Here o = Bin x, do = cob xdx, and n = 2, 4, and 6 respectively. 

Ex. 2. Find rco8*xdx. 

SoXvXvm. fcoB^xdx = j cos^x cos xdc = Ml — sin^x) cos xdx 

= jcos xdx — r Bin*x cob xdx 

sin^x . ^ 
= sin X — V C. 

EXAMPLES 

//* sin'x 
sin*x(2x = \ co8*x — cos x + 0. 2. I sin^x cob xdx = 1- C. 

3. iBin^xdx = — cos x + - cob*x \- C. 

J 3 5 

4. fcoB^xdx = sin X — - sin'x H h C. 

5. I Bin « COB 808 = V C. 

4/ 2 

6. r coB^x sin'xdx = — J cosPx + } cos'^x + C. 

7. ( COS*a sin ada = ^— ^ + C. 

•/ 3 

Q fcos^xdx 1 « . /^ o rsin'oda . . ^ 

8. 1—7-: — = C8Cx--CBC«x + C. 9. I — = 8eca + C06a + C. 

•/ sin^x 3 J cos^a 

10. fsin^^ coB»0d^ = ^ sin^0 - J^ sin^^ + C. 

11. fsin'^ co8«^d^ = { sin*d - ^^ am^e + V*? sin^*^ + C. 



INTEGRATION 805 

12. f-^^dif = - 2 Vcos y (1 - ? coeay + -co8*y) + C. 

13. r^?^ = ?8inlt(l-lBin8t + -8in<t) + C. 

V5^ 2 2 7 

Example II. To find i tan* xdx, or I coV xdx. 

These forms can be readily integrated, when n is an integer, on 
somewhat the same plan as the previous examples. 

Ex. 1. Find ftan* xdx. 

BoLviwn, ftan* xdx = ftan^x (sec^ x - 1) dx 28, p. 2 

= ftan^ X 8ec2 xdx — jtan* xdx 
= r(tan x)'cl(tan x) - ("(sec^x - l)(fe 
= — tan X + X + C. 

Example III. To find j 9e(f xdx^ or J cs(f xdx. 

These can be easily integrated when n is an even positive integer. 

Ex. 2. Find jsec^xdx. 

SohUUm. fsec^ xdx = r(tan«x + 1)* 8ec« xdx 28, p. 2 

= Mtan x)* sec* xdx + 2 ntan x)' sec* xdx + Jsec* xdx 

= h2 htanx + C. 

5 8 

Example IV. To find ( tan'^x %e<f xdxj or i coi^x cs(f xdx. 
When n is a positive even integer we proceed as in Example III. 

Ex. 3. Find jtan^xsec^xcEx. 

SoMUm, fton^x Bec*xdx = ftan^x (tan^x + 1) sec^xdx 28, p. 2 

= Otan x)* sec^xdx + Jtan^x aec'xdx 

^tan^ tanTx^^ By [4] 

9 7 

Here « = tanx, do = sec* xdx, etc. 



806 INTEGRAL CALCULUS 

When m is odd we may proceed as in the following example. 
Ex. 4. Find J tan^x sec'xda; = } tan^x sec'x sec x tan xdx 

= r(sec*x - 1)« sec^x sec x tan xdx 28, p. 2 

= Hsec'x — 2 sec*x + sec«x) sec X tanxdsD 
sec'^x 28ec*x . sec^x ^ _ 

Here o = 8ecx, do = sec x tan xdx, etc. 

EXAMPLES 

1. rtan»xdx = — ^- + logcosx + C. 
J 2 

2. J'cot«xdx = -?^-log8inx + C. 

3. fcot^^dx = ~ cot»5 + 3 cot? + X + C. 
•^8 S3 

4. Jcot*xdx = -cotx-x + C. 

6. fcot^oda = - } cot*« + i cot«o + log sin a + C. 

a ftan»^dy = tan*^ - 4 tan« 7 + 4 logsec ? + C. 
•^ 4 4 4 4 

!• r . J tan^x 3tan'x 

7. I 8ec»xdx = — — + ^^=^ + tan»x + tanx + a 
•^ 7 6 

8. Jcsc^xdx = — cot X — } cot^x — \ cot* x + C 

9. rtan«*Bec«^ = *55l* + *5B!^ + c. 
•^ 7 6 

10. rtan*08ec*9({0 = |8ec'0 - J 8ec>0 + C. 
12. Jtanlxi«c4«te = !l!^ + 2tan»«^p 

14. P^ = tan. -2cot.- £2^ + 0. 
J tan*a 8 



INTEGRATION 307 

15. r(tan«z + tan*2)dz = ^tan»2 + C. 

16. ("(tan t + cot «)8 (ft = J (tan« t - cot* <) + log tan^ t + C, 

Example V. To find isinTxcoif'xdx by means of multiple angles. 

When either m orn is a positive odd integer the shortest method 
is that shown in Example I, p. 303. When m and n are both 
positive even integers the given differential expression may be trans- 
formed by suitable trigonometric substitutions into an expression 
involving sines and cosines of multiple angles, and then integrated. 
For this purpose we employ the following formulas : 

sin ti cos u = i^ sin 2 u, 86, p. 2 

sin*w = i — ^ cos 2 w, 38, p. 3 

cos^te = ^ + ^ cos 2 u. 39, p. 3 

Ex. 1. Find fcos^xdx. 

Solutum. Cco^xdx = f (i + J cos 2 x) dx 38, p. 3 

= i fdz + - rco82«te=:- + 78in2x+C. 
2J 2J 2 4 

Ex. 2. Find fsinaxcos^xdx. 

Solution, fsin^xcosaxdx = J rflin22xdx 36, p. 2 

= if a - i cos4x)(fe 38, p. 3 

= sin 4 X + C 

8 32 

Ex. 3. Find fsin^xcos^xcte. 

Solution, r sin^ x cos^ xdx = j (sin x cos x)' sin^ xdx 

= rjsin22x(i-icos2x)dx 36, p. 2 ;. 38, p. 3 
= J rsin32xdx - J rBin2 2xco8 2xdx 

= if a - ico8 4x)dx - J rsin«2xcos2x(ix 

_ X sin 4 X sin' 2x ^ 
~ 16 64 48" 



808 INTEGRAL CALCULUS 

Example VI. To find i sin mx cos nxdxj Csin mx sin nxdzj or 
Ccos mx cos nxdxj when m ^ n. 

By 41, p. 8, smmxcosnx = ^8iii(m + n)x-\'^Bm{m'-'n)x. 

.'. j sin mx cos nxdx = J^ rsin {m -f- n) scdx + J j sin (w — n) xdx 

_ cos(^+yt)^ C0B{m — n)x ^ 
^ 2(w + n) 2{m — n) 

Similarly we find 

/, 8in(wi-|-n)a: . sin (m — n)a; . ^ 
sm mx sin nxdx = rr^ — ■ — ^ H — ^ '— + C, 
2 (w -H n) 2 (m — n) 

J, 8in(7n-f-n)a: sin(m — n)a; . ^ 
cos WW? cos nxdx = — ;r-^ — ^ — ^— + -—-^ ^ + C 
2(?n + n) ^ 2(w-n) 

EXAMPLES 

/x 1 
co8«xdx = - + -8in2x + C. 
2 4 



2 



3x sin 22 . sin 42 



4. Isin^xdx = — ( 52 — 4 8in22 + — h-Bin42) + C 

•/ 16 \ 3 4/ 

6*. rco8«2d2 = — (52 + 48in22 — -? + 78in42 ) + C. 

•/ 16 \ 3 4/ 

y. r . -1 o J 8in«2a .a 8ln4a _ 

6. I Bin* a C08» odo = -— + + C. 

J 48 16 64 

7 rsm*«coB«W< = — (3«-8in4« + ^^^) + C. 
•/ 128 V 8 / 

8. rc(M^2 8in*2d2 = — (62 H- -8in«22 — 8in42 l + C 

J . 128 V 3 8 / 

A r o • e J cosSy C082y . _ 

9. I COB 3y sin 5^(2^ = + C 

•i/\ r • c • a J sinllz sinz . ^ 

10. I Bin 6 z Bin 6 zdz = \- h C. 

J 22 2 

•■I r .1 ^ J sinlls . 8in3« . ^ 

11. I cos 4 « cos 7 ad« = h C 

J 22 6 

12. j COS} 2 sin} 2(22 = — I cos 2 + 006^2 + C- 

13. Jcos 3 2 COS I xdx = ^\ sin Y 2 + A 8in J 2 + C. 



CHAPTER XXV 
COHSTAHT OF IHTXORAIIOII 

185. Determination of the constant of integration hy means of 
initial conditions. As was pointed out on p. 290, the constant 
of integration may be found in any given case when we know the 
value of the integral for some value of the variable. In fact it is 
necessary, in order to be able to determine the constant of inte- 
gration, to have some data given in addition to the differential 

, expression to be integrated. Let us illustrate this by means of 
an example. 

Ex. 1. I^ad a. fonctioti whose flrat derivatiTs is SsC — 2z + 6, and which shall 
hare the ralne 12 when z = 1. 

Solution. (3 x^ — 2 z -)- 6) de 18 the diSerential expreaalou to be Integrated. Thus, 
r(3i» - 2i + 6)<it = x» - z» + 6* + C, 

where C is the constant of integration. From the conditlone of our problem this 
nsolt must equal 12 when z = 1 ; that is, 

12 = 1-1 + 6 + 0, or, C = 7. 
Hence z*— z* + 6x + 7l8the required function. 

186. Geometrical signification of tlie constant of integration. We 
shall illustrate this by means of examples. 

Ex. 1. Determine the equation of the curve at 
every point of which the tangent has the slope 2z. 

SobitUm. Since the slope of the tangent to a curve 
at any point Is — , we have by hypothesis 
*** 

dx 
or, in = 2xdx. 

Intagratlng, j/ = ifxdx,OT, 

(A) K = *'+C, 

where C is the constant of integration. Now if we give 
a series of values, say 0, 0, — 3, (A) yields ibe 



equations 



V = z» + 6, i/ = x\ v = 



810 INTEGRAL CALCULUS 

irhoge loci are parabolas with axes coinciding with the sxia of y and having 6, 0, — S 
respecilvelf as inlercepte on the axis of Y. 

All of the parabolas (A) (there are an Infinite number of them) have the same 
value of — ; that is, they have the same direction (or slope) for the same value of X. 
It vrill also tie noticed that the difference In the lengths of their ordinates remains 
the same for all values of z. Hence all the parabolas can be obtained b; moving 
any one of them vertically up or down, the valae of C in this case not affecting the 
slope of the carve. 

If In the above example we impose the additional condition that the curve shall 
pass through the point (1, 4), then the coBrdinates of this point most satis:^ {A), 

*"°« 4.1 + C, or, C = 8. 

Hence the partlcnlat curve required Is the parabola y = a* + 3. 

Ex. 2. Determine the equation of a curve such that the slope of the tangent to 
the curve at any point Is the negative ratio of the abscissa to the ordinate. 

Solulion. The condition of the problem is ez* 
pressed by the equation • 



or, separating the variables, 

ydii = '-zdx. 

Integrating. ^ = -_ + c, 

or, I* + v« = 2 C. 

Tliis we see represents a series of concentric circlM 
with their centers at the origin. 
If, in addition, we impose the condition that the curve must pass through the 
point (3, 4), lb.n » + 16-20. 

Hence the particular curve required Is the circle x* + ^ = 26. 

187. Physical signification of the constant of integration. The 
following examples will illustrate what is meant. 



s In a ett^ght 



Solution. Since the acceleration \=— from (U), p. lOfil ia constant, say/, 
h«vB L dt J 



dt ■^' 
ir, do =/d(. Integrating, 

(.i) e =^ + C. 

To delwmlne C, suppose that the initial velocity be «( 
r = Co when f = 0. 



CONSTANT OF INTEGRATION 811 

These values substituted in {A) give 

to = + C, or, C = Vq. 
Hence (A) becomes 

(B) «=yi + «o. 

Since » = ;^ Fw, P. lOs] , we get from (B) 

or, da=ftdt + V(4t. Integrating, 

(C) «=iyr»-f ro« + C. 

To determine C, suppose that the initial space (= distance) be «o; that is, let 

s = So when t = 0. 

These values substituted in (C) give 

«o = -f + C, or, C = So. 
£ence (C) becomes 

(D) » = i/P + »o« + »o. 

By substituting the values f=g, Oo = 0, So = 0, s = ft in (£) and (D) we get the 
laws of motion of a body falling from rest in a vacuum, namely, ' 

(Ba) v — gt^ and 

(Da) A = ip^. 

Eliminating t between (Ba) and (DcC) gives 

Ex. 2. Discuss the motion of a projectile having an initial velocity vo inclined 
an angle a with the horizontal, the resistance of the air being neglected. 

SolvJtUm. Assume the JTF plane as the plane of motion, OX as horizontal, and 
OY as vertical, and let the projectile be thrown from the origin. 

Suppose the projectile to be acted upon by 
gravity alone. Then the acceleration in the hori- 
zontal direction will be zero and in the vertical 
direction — p. Hence from (15), p. 106, 

Integrating, Vx = C\ and Vy = — gt-\-C%. 

But oo cosa = initial velocity in the horizontal direction, 

and Vosin a = initial velocity in the vertical direction. 

Hence C\ = Vo cos a and d = Vo ^^ a, giving 

(E) «x = »o cos a and Uy = — gft -f «o sin «• 

But from (10) and (11), p. 104, »x = — and »if = -^ ; therefore (E) gives 

dx J dy ^ , ' 

— =Vo cos a and — - = — fl« + to sin a, 
dt at 

or, dz= VoCOBadt and dy=z — gtdt + i^o Bin adt. 




312 INTEGRAL CALCULUS 

Integrating, we get 

(F) X = »o cos tt • t + C» and y =:— \gfi + VoBma-t + Ci, 

To determine C3 and O4, we observe that when 

t = 0, jc = and y = 0. 

Substituting these values in (F) gives 

Cs = and C4 = 0. 
Hence 

(O) x = vo cos a • t, and 

(H) y = —\ffi^ + Vo^na't, 

Eliminating t between {O) and (H), we obtain 



(7) y = X tan a — 



gx^ 



2ro*co82a 

which is the equation of the trqjectoryj and shows that the projectile will move in 
a parabola. 

EXAMPLES 

Find the function whose first derivative is 

x' 

1. X — 3, knowing that the function equals when x = 2. Ans, -— — 3 x + 18. 

2. 3 + X — 6 x^, knowing that the function equals — 20 when x = 6. , . . 

Ans, 804 + 3x + |--^- 

2 8 

3. (y* — &^), knowing that the function equals when y = 2. 

4 2 

4. sin a + cos a, knowing that the function equals 2 when a = -• 

Ans, sina — cosa + l. 

6. 1 knowing that the function equals when t = 1. Ana. log (2 1 — t"). 

Find the equation of a curve such that the slope of the tangent at any point is 

6. 3x-2. Ans. y = — -2x + C. 

2 

7. xy. Am. y = ce*. 

8. x2 + 6x, the curve passing through the point (0, 3). ^ ^ 

^iw. y = i. + -^ + 3. 

9. -f the curve passmg through the point (0, 0). Ans, y> = 2px. 

y 

10. — , the curve passing through the point (a, 0). Ans, 6«x* — a^ = a^. 

11. m, the curve making ^n intercept b on the axis of y. ^n«. y = mx + b. 



CONSTANT OF INTEGRATION 818 

Find the relation between z and y, knowing that 

12. *? = ^. Ana. ^ = ^-^0. 

dx y 2 3 

' 13. xdy -f ydx = 0. Aiis, xy = C. 

14. ^ = 1±-?, If y = when x = 0. Ana. x« + y« + 2jc - 2y = 0. 

dz 1 — y 

16. (l-y)da;-f (l + x)dy = 0. Ana, logl±^ = C. 

16. Find the equation of the cuire whose subnormal is constant and equal to 2 a. 

^j. Ana. y^ = 4 ox + C, a parabola. 

JTiAl. From (4), p. 90, subnormal ^y—' 

dx 

17. Find the curve whose subtangent is constant and equal to a [see (3), p. 90]. 

Ana. a logy = X + C 

18. Find the curve whose subnormal equals the abscissa of the point of contact. 

Ana. y^ — x^ = 2 C, an equilateral hyperbola. 

19. Find the curve whose normal is constant (= iZ), sssuming that y = jR when 
x = 0. Ana. x« + y« = iP, a circle. 

Biitt. From (6), p. 90, length of normal « y V 1 + (-^ j i or, dx = * ( JP - y^" ^ydy. 

20. Find the curve whose subtangent equals three times the abscissa of the 
point of contact. Ana. x = cy^. 

21. Show that the curve whose polar subtangent [see (7), p. 99] is constant is 
the reciprocal spiral. 

22. Show that the curve whose polar subnormal [see (8), p. 99] is constant is 
the spiral of Archimedes. 

23. Find the curve in which the polar subnormal is proportional to the length 
of the radius vector. Ana. p = ce°^. 

24. Find the curve in which the polar subnormal is proportional to the sine of 
the vectorial angle. Ana. p = c—a cos 6. 

26. Find the curve in which the polar subtangent Is proportional to the length 
of the radius vector. Ana. p = ce^. 

26. Determine the curve in which the polar subtangent and the polar subnormal 
are in a constant ratio. Ana. p — ctf^. 

27. Find the equation of the curve in which the angle between the radius vector 
and the tangent is one half the vectorial angle. Ana. p = c (1 — cos B). 

Assuming that v = Vo when £ = 0, find the relation between o and £, knowing 
that the acceleration is 

28. Zero. Ana. o = 0o. 

29. Constant = k. Ana. v — VQ + ld. 

30. a + W. Ana. t = t>o-f crf4- — • 



814 INTEGRAL CALCULUS 

Assuming that s = when t = 0, find the zelation between 8 and t, knowing 
that the velocity is 

31. Constant (=«o). -^w- « = »ot. 

32. m + iU. -Arw. « = m<-f — . 

33. 3 + 2«-3t». ^»w. « = 3«-ft*-(\ 

34. The velocity of a body starting from rest is 6 {* feet per second after t sec- 
onds, (a) How far will it be from the point of starting in 3 seconds ? (b) In what 
time will it pass over a distance of 360 feet meaflured from the starting point ? 

Ans, (a) 45 ft.; (b) 6 seconds. 

36. A train starting from a station has after t hours a speed of ^ — 21 1^ + 30 1 
miles per hour. Find (a) its distance from the station ; (b) during what interval 
the train was moving backwards ; (c) when the train repassed the station ; (d) the 
distance the train had traveled when it passed the station the last time. 

An8. (a) 1 1« - 7 1> + 40 <3 miles ; (b) from 6th to 16th hour ; 
(c) in 8 and 20 hours ; (d) 4658^ miles. 

36. A body starts from O and in t seconds its velocity in 
the X direction is 12 e and in the Y direction 4<s - 0. Find 
(a) the distances traversed parallel to each axis; (b) the dis- 
tance traversed along the path ; (c) the equation of the path. 

An8, (a) x = 6f^ y = -<«-9t; 




(b), = lt« + 9t; (c)y = (^x-9)^|. 



37. The equation giving the strength of the current i for the time t after the 
source of E.M.F. is removed, is (R and L being constants) 

Find i, assuming that I = current when t = 0. Ans. i = le l, 

38. Find the current of discharge i from a condenser of capacity C in a circuit 
of resistance £, assuming the initial current to be Joi having given the relation 

di_ dt^ 

i " CjB' j_ 

C and R being constants. Ans. % = Jo^ci?. 



CHAPTER XXVI 

INTEGRATION OF RATIONAL FRACTIONS 

188. Introduction. A rational fraction is a fraction the numerator 
and denominator of which are integral rational functions.* If the 
degree of the numerator is equal to or greater than that of the 
denominator, the fraction may be reduced to a mixed quantity by 
dividing the numerator by the denominator. For example, 

= ar-f a; — 3 + 



x^-\.2x-^l ar^-f2a:-fl 

The last term is a fraction reduced to its lowest terms, having 
the degree of the numerator less than that of the denominator. 
It readily appears that the other terms are at once integrable, and 
hence we need consider only the fraction. 

In order to integrate a differential expression involving such a 
fraction it is often necessary to resolve it into simpler partial frac- 
tions, i.e. to replace it by the algebraic sum of fractions of forms 
such that we can complete the integration. That this is always 
possible when the denominator can be broken up into its real prime 
factors will be shown in the next section.f 

189. Partial fractions. Consider the rational fraction 

(A) ^^ 

reduced to its lowest terms, the degree of the numerator being less 
than that of the denominator. If a occurs a times as a root of the 
equation /(a;) = 0, we may write 

* That to, the Tariable is not aifected with fractional or negative exponents (see $25, p. 16). 
t Theoretically, the resolution of the denominator into real quadratic and linear factors is 
always possible when the ooeffl<;ients are real, that is, such a resolution exists. 

816 

J 



816 INTEGRAL CALCULUS 

where <^ {x) is not divisible hj x — a. Therefore <^ (a) ¥= since a is 
not a root of <^ (2;) == 0. We may then write {A) in the form 

(5) ^<^> 



(a;— aY^{x) 
The equation 

F{x) ^ A F{x) A 

f{x) (a; — a)* (a;— a)«<^(a;) (a; — a)* 

is evidently true for any value of A, 
Combining the last two fractions, 

(0) J^(x)^ A F{x)-Ai>{x) 

^ ' fix) {x-aY (x--a)-<t>{x) 

Now let us determine the value of A so that the numerator of 
the last fraction in ((7) shall be divisible by x — a. In that case 
x=za will be a root of F(x) — A<l> (x) = 0, and hence 

F{a)^A<l>{a)=0, or, A = ^. 

This value of A is finite since <f> (a) =^ 0. 

Having now determined the value of A so that x — a shall be a 
factor of F{x) — A^{x)^ we may write 

F{x) - A4>{x) = {x - a)F^{x), 

where the degree of the new function F^ {x) is less than the degree 
of (a:-a)«-*<^(a:). 
Hence from {C) 



f{x) {x-aY {x-aY-'4>(x) 

If a > 1, we proceed in the same manner with the second fraction 
on the right-hand side of (Z)), giving, say, 

where A. = ^^ . 

<^(a) 



. mXEGRATION OF RATIONAL FRACTIONS 817 

Continuing this process, we get 

^ ' /(x) {x-a)'^{x-a)'-'^ ^x-a^<l,{x) 

where G (x) is of lower degree than <t> (x), and <f> (x) does not contain 
the factor x — a. 

If now b OCCUI8 l3 times as a root of f{x) = 0, it will also occur 
13 times as a root of ^(x)= 0, and we may break up 

GJx) 
<l>{x) 

into a sum of partial fractions in exactly the same way as we 
decomposed the original fraction. 

From the preceding discussion it follows that if a, J, <?,.., Z are 
roots of the equation /(a;) = 0, occurring respectively a, /8, 7, ..., X 
times, the given rational fraction may be broken up into a sum of 
rational fractions as follows : 



w S=7A-.+r-^+-+^ 



a-1 



f{x) (x — a)' (z — a)*"' x—a 

+ ^ + ?l + ... + ^£=1 

{x-bf (x-bf-^ x-b 

"^ (a; _ Z)* "^ (X - i)*-' + ■ • • + ^37' 

Since every equation of degree n has n roots, it is evident that 
there will be as many partial fractions as there are units in the 
degree ot f(x). 

Instead of finding the constants A^A^^ • • •, J9, ^^ . . ., Z, Z^ --- by 
the method indicated above, it is more usual to clear ((7) of frac- 
tions. This makes F(z) identically equal to a polynomial in x of 
degree not greater than n — 1 [assuming n as the degree of F{x)] 
and therefore containing at most n terms. Since this identity 
holds true for all values of Xj we equate the constant terms and the 
coefficients of like powers of x on both sides. This gives n inde- 
pendent and consistent equations, linear in the constants required. 
Solving these n simultaneous equations we get the n constants 



318 INTEGRAL CALCULUS 

190. Imaginary roots. Equation ((7) on p. 317 holds true when 
some or all of the roots a, i, c, • - , 2 are complex (see § 11, p. 9). 
When the coefficients of F{x) and f{x) are real — and this is the 
only case we shall consider — we may avoid the complex numbers 
and put our results in real form as follows. 

Suppose, for instance, that the root b is complex. It may be 
written in the form j^^^;^,, 

where g and A are real numbers and i = V— 1. Then there must 
be a second complex root, say c, conjugate to (, namely, 

<? = ^ — hi. 

If h occurs fi times and c occurs 7 times, we see that fi must 
equal 7. From the manner in which J5, B^^ • • •, C, Cj, . . . were deter- 
mined it is evident that if 



J?= 6? -h Hi, B^ = G^ 4. HJ,, 
we shall have 

C^G-Hi, C^ = G^^H^i, 

Hence the sum 



'•» 



a a . . • 

x—h X— c 

may be expressed in the real form 
(B) t(£) , 

where the denominator is of degree 2/8 and the numerator of 
degree not greater than 2/8 — 1. Let '^i(:r) be the quotient and 
P^x-^-Q^ the remainder found in dividing '^{x) by [(x — ^)*-^ ^*]a 
Then ^ (^) ^ |-(^ _ gy ^ ^r| ^^ (^) ^ p^^ ^ ^^^ 

and {B) may be written in the form 

IC\ Vr(a:) ^ P^x + Q, ^,{x) 

^ [(^-9f + hy [{x-9Y + h'Y^[{x-gY + h'Y-'' 



INTEGRATION OF RATIONAL FRACTIONS 819 

Similarly the second fraction on the right-hand side of ((7) may 
be written in the form 

ilh V^i(^) ^ A^-hg, JTti^) 

Continuing this process, (A) gives finally 

^ ' (x-bf'^ {x-bf-^'^""^ x-b 

+ — ^—-\- ^ + --- + -^^ 

{x — cy {x — c)^^ x—c 

\{,x^gf^h^^^\{x-gfJr't^'Y-" ix^gf + h^' 

The coefficients P^ Q^, P„ Q^ . • ., P^, Q^ are calculated by the 
same method we employed to calculate the coefficients in the last 
section. 

We shall now proceed to illustrate what has been said by work- 
ing out numerous examples in detail. It is convenient to classify 
our problems imder the following four heads. 

Case I. When the roots of f{x) = are all real aod none 
repeated. The denominator may then be broken up into real 
linear factors, none of which are repeated. 

Case II. When the roots of f{x) = are all real but some 
repeated. Then the denominator may be broken up into real 
linear factors, some of which are repeated. 

Case III. When the equation /(a;) = has some imaginary roots, 
none of which are repeated. Then the denominator may be broken 
up into a product of real prime factors, there being a real quad- 
ratic factor (factor of the second degree) corresponding to each 
pair of conjugate imaginary roots. 

Case IV. When the equation f{x) = has some imaginary roots 
repeated. The corresponding quadratic factors are then repeated 
in the denominator. 

When we speak of factors of the denominator we shall mean only 
real prime factors, as these include all the types that can occur.* 

* tf the ooefflcient of the highest power of the rariahle in the denominator is different from 
nnity, we begin by dividing both numerator and denominator by this coefficient. 



820 INTEGRAL CALCULUS 

191. Case I. When the factors of the denominator are all of the 
first degree and none repeated. 

It is evident from ((y), p. 317, that to each non-repeated linear 
factor, such as 2; — a, there corresponds a partial fraction of the form 

A 

X— a 

Such a partial fraction may be integrated at once as follows : 

•/ x-^ a %/ X — a 
Ex. 1. Find C^l±3^. 

Solution, The factors of the denominator being x, x — 1, x + 2, we ajBSume* 

2x + 3 A , B , C 

W Z7Z TTTII-r^ = T + 



x(x-l)(x-f2) X x-1 x + 2 

where Aj B, C are constants to be determined. 
Clearing {A) of fractions, we get 

{£) ' 2x + 8=^(x-l)(x + 2)+B(x + 2)x + C{x-l)x 

= (A + B 4- C)x2 + (^ + 2 B - C)x - 2^. 

Since this equation is an identity, we equate the coeiBcients of the like powers 
of X in the two members according to the method of Undetermined Coefficients, 
and obtain the three simultaneous equations 

(C) Ja + 2B-C = 2, . 

l -2^ = 3. 

Solving equations (C), we get 

^=~f, B=J, C = -f 
Substituting these values in (A), 

2x + 3 __JL+ ^ ^ 



x(x-l)(x + 2) 2x 3(x~l) 6(x + 2) 

^x(x-l)(x + 2) 2^ X 3^x-l 6^x4-2 

= - } log*c + f log(x - 1) - i log(x + 2) + logc 

, c(x-l)« . 

x*(x + 2)* 

A shorter method of finding the values of ^, B, and C from (B) is the following: 

Let factor x = ; then 3 = ~ 2A^ or, ^ = — {. 

Let factor x — 1=0, orx = l; then 5 = 3 B, or, B = }. 

Let factor x + 2 = 0, orx = -2; then ~ 1 = 6 C, or, C = - J. 

*In the proc;e88 of decomposing the fractional part of the given differential neither the 
Integral sign nor dx enters. 



INTEGRATION OF RATIONAL FRACTIONS 821 



^ r (x-l)dx ^ J c(z -f4)t 
' J x2 + 6x + 8 (x4.2)»* 

J x»-4x 3 2 * (x + 2)« 

. f x*dx X» _ 1, x-1 16, , . ov . r» 

6. r <''-'')y''y iog('^-'')%c. 

/• (!P + pq)S U-p)(t + q) 

'• J t(rri,)>T^ - >°« i + ^• 

v^z*-5«« + 6 2V2 Z+V2 2V3 « + V3 

192. Case II. When the factors of the denominator are aU of the 
first degree and some repeated. 

From (G)y p. 317, it is clear that to every Wrfold linear factor, 
such as (2; — a)", there correspond the n partial fractions 



(x — a)" (x — a)*^ x^a 

The last one is integrated as in Case I. The rest are all inte- 
grated by means of the power formula. Thus, 

J{x-ay J^ ^ (1 - w) (a; - a)-^ 



x« + l 



Ex. 1. Find f f ,,. dx 



SolutUm. Since x — 1 occurs three times as a factor, we assume 

x» + l ^ . B . C . D 

= r ■ 771 "T Z 77^ "" 



x(x-l)» X (x-l)» (x-l)« x-1 
Clearing of fractions, 

x« + 1 = il (X - 1)» + Bx + Cx(x - 1) + I>x{x - 1)« 

=z{A +2))x« + (-8A + C-22))x2 + (3-4 + B-C + D)x--4. 



822 INTEGRAL CALCULUS 

EqaatiDg the coefficients of like powers of x, we get the simultaneoas equations 

-4 + D = l, 

-3A + C-2D = 0, 

3^ + B-C + l>=0, 

-^ = L 

Solving, -4 = - 1, B = 2, C = 1, D = 2, and 

x» + l 1. 2 , 1 . 2 



x(a;-l)« X (x-l)» (x-l)« x-1 



(X-l)2 ° X 



EXAMPLES 



1. f -^ = _±_4.iog(x + l) + C. 

J (X + 2y(x + 1) x + 2*^ ' 

(X - 1)1 (» - 1)^ X - 1 * ^ ' 

^ r(3xj,2)<ix^jlx4j_ _x^ 

J x(x + l)<» 2(x + l)a ^(x + l)a 

5 r g'dx _ 6x + 12 /x4-4y _ 

* J (X + 2)2(x + 4)4 " x^-f 6x + 8"^ ^^\x + 2/ "^ ' 

6. r ^^ = — ^ + log(y + 1) + C. 

•J(t^-2)2 4(^2-2) '^8V2''^t-V2 ' 



cui^ , . V . 2a4 



8. I — -— = a log(« + a) H + C 

J (a-|-a)» *^ ^ « + o 2(a + a)2^ 

^ \z + m (z + n)2/ ^'^ z + n 

193. Case III. When the denominator contains factors of the 
second degree hut none repeated. 

From (E)^ p. 319, we know that to every non-repeated quadratic 
factor, such as a;*+jt>a;-f g, there corresponds a partial fraction of 
tiie form ^:r + ^ 

3?J^px^q 



INTEGRATION OF RATIONAL FRACTIONS 323 

This may be integrated as follows : 



/ 



4 



(Ax + B) dx _ I V 2 2 y 




Ax + ^-^ + B"\dx 

Adding and subtracting — ^ in the numerator. I 

-f + B^dx 




— Ap\ /• dx 



_ A r (2x-\^p)dx / 2B 
'"2*/a?-f-jt>a? + 5' \ 




i)H'-^) 



[Completing tlie square in the denominator of the second integral.] 
^ ^ /J. .v.2^— ^O ^ 2x-|-» ^ 

= TT ^og{ar-\-px + q)-i — ^ arc tan — Z= ^= -h C. 

Since :if -\r px -if q^^ has imaginary roots, we know from 3, p. 1, 
that4g-./>0- 



X 



Ex. 1. Find f-r^ 

Solution. Aflsame — = — | — 

x(x2 4.4) X x2 + 4 

Clearing of fractions, 4 =-4 (x« 4- 4) + x(Bx+ C)= (-4 + B)x2 + Cx + 4A. 
Equating the coefficients of like powers of x, we get 

-4 + 5 = 0, C = 0, 4^=4. 

4 1 X 

This gives A = 1, B = - 1, C = 0, so that = — 

* » » » x(x2 + 4) X x2 + 4 

/4 dx __ rdx ^ r zdx 
x(xa + 4)"J 'x"'~Jx« + 4 

= logx — - log(x2 + 4) + log c = log ^ . il?w. 

2 VxM-4 

EXAMPLES 

, r «lx 1 , x« + 4 2 ^ X . ,7 

1. I = — log 4- - arc tan - + C. 

J (X + 1) (X» + 4) 10 * (X + 1)8 6 2 

^ r (2x«-3x-8)dx , (x«-2x + 6)* 1 ^ x-1,^ 

2. I -^ = log ^ + - arc tan \- C. 

J (x-l)(xa-2x + 6) ^ x-1 2 2 



„ r x«dx 1 , 1 + X 1 ^ . ^ 

3. I = - log arc tan X + C 

J 1-x* 4 ^1-x 2 



824 INTEGRAL CALCULUS 

. r dx 1, aj* 1 . _ 

4. I — = T log = arc tan x + C. 

5- I -4 — TT-i — :: = log , + - arc tan arc tan — — + C. 

r (6x' — l)dx , X*— 2x4-6 5 x—l 2 x ^ 

6. I — ^ = log — h -arc tan -arctan^-+C 

J (xa+3)(x2-2x+6) *x2 + 32 2 Vs Vs 

^ f dx 1. (x + 1)" 1 , 2x-l , ^ 

7. I — = - log -^ h — - arc tan — - — \- C. 

Jx«-4-l 6 *x8-x+l Vi V3 

8. I — ^ — = log + - arc tan - + C. 

Jx« + x2 + 4x4-4 *(x + l)^ 2 2 

9. I = - log I 1 + -z- arc tan — - + C 

10. I = — - log — ;= h V 2 arc tan h C 

,, r dy 1, i^ + y + 1 . 1 ^ 2y4-l . ^ 

11. I __*-- = - log -| — ^ r + -^arctan ^ + C. 

194. Case IV. TF%6n the denominator contains factors of the 
second degree some of which are repeated. 

To every w-fold quadratic factor, such as (a?-fj9a;+g)", there 
correspond the n partial fractions 

/J\ ^^4-^ . Cx + I) Lx^M 



To derive a formula for integrating the first one we proceecl 
as follows : 



/; 



Ax + f-f + B^dx 
{a?+px + q)' "" ~J {3»+px + qy 






^ + S . . _ V " 2 2^ J 




I Adding and subtracting -^ in the numerator. I 

Ax + ^^dx C(-^ + B\dx 






(7?^px+qY J {3? + pxJrqY 
d C,-^ ^ „. ^ .^-./o . ^ „x J. ^ /2B-Ap\ C dx 



J(.. +;,x + ,r(2 - +l>)ix + (i«^)/^^ ^^^ ^ ^^. 



INTEGRATION OF RATIONAL FRACTIONS 826 

The first one of these may be integrated by [4] ; hence 



(^ /i 



Ax-^-B 



dz = 



{J? +px + qY 2 (1 - n) (a? J^px + qf'^ 

dx 



+ 



(^^)/; 



{a?^px^qy 



hiL. 



Let us now dijBEerentiate the function -,, 

{^ ■¥ px -\' qy-"^ 

Thus, 

- + | \ 1 2(n-D(.+f 



S 



dx \{2^'\'px -f g)*"7 {^ -i-px + q)*~^ {3? +px + qy 



(0) . d 



^2 



{^'¥pX'\'qT-\ 



{3?+px + q)'-^ (sd'+px + q)' / 

fsiaoe 3fl+px+q~(x+^+(q-^, «nd («+^ -(«'+J»*+«)-(«-^-l 

Integrating both sides of (<7), 

P 

« + *- 

2 .„„./» da; 






+ 2(„-l)(,-f)/- 



(fi; 



or, solving for the last integral, 

J{^+px^qr 2in-^l)(q^fji:^^px-^qr^ 



-f 



or. 



2n-3 r dx 



326 INTEGRAL CALCULUS 



Substituting this result in the second member of (5), we get* 

{Ax + B)dic _ A(p*^4:q) + {2B^Ap)(23D+p) 
(ic*+px + q)^ 2(n — l)(4g —!>■){»■+ pa? + gr-* 

die 



mf; 



(2B - Ap) (2n - 3) r 

"*■ (n-l)(4g-i>*) J {X* + px + q)*^-^ 

It is seen that our integral has been made to depend on the 
integration of a rational fraction of the same type in which, how- 
ever, the quadratic factor occurs only n — 1 times. By applying 
the formula (E) n — 1 times successively it is evident that our 
integral may be made ultimately to depend on 

dx 



f-. 



0^ -\-px-\-q 

and this may be integrated by completing the square, as shown 
on p. 802. 

In the same manner all but the last fraction of {A) may be 
integrated. But this last fraction, namely,* 

Lx-\'M 

—5 1 

or '{'px-\'q 

may be integrated by the method already given under the previous 
case, p. 823. 



Ex. 1. Find 



/ 



(a;« + a;« + 2)(iaj 



{x2 + 2)2 

Sohdion, Since 2^ + 2 occurs twice as a factx)r, we assume 

g» -i- x^ + 2 _ Az-\- B Cx-\- D 
(x2 + 2)2 " (x2 + 2)2 "^ xs + 2 * 
Clearing of fractions, we get 

x» + x2 + 2 = ^x + B + (Cx + D) (x2 + 2) 

= Cx8 + Dx2 + (^ + 2 C)x + B + 22X 

Equating the coefficients of like powers of x, 

C = l, D = l, ^ + 2C = 0, B+2D = 2. 

This gives ^ = - 2, B = 0, C = 1, D = 1. 

„ x8 + a;2 + 2 2x x + 1 

Hence — ^^— = 1 — — — , and 

(x3 + 2)2 (x2 + 2)2^x2 + 2 



/ (xg + x2-i-2)dz _ _ r 2zdx r xdx r dz 
{X2 + 2)2 "J (X2 + 2)2 "^ J X2 + 2 "^ J X2^ 



2 



1 X 1 

+ -— arctan— - + -log(x2 + 2) + C 



«^ + 2 ^^ \/2 2 

• 4g-/>«>0, since x*-^px + q=0 has imaginary roots. 



INTEGRATION OF RATIONAL FRACTIONS 827 

Ex. 2. Find fg^ + ^ + Sdx. 

Solution. Since x^ _|. i occurs twice as a factor, we assume 

2x* + x + S _ AX'{-B Cx + D 
{x«+l)a ~(x2 + i)a'*" a^ + i * 

Clearing of fractions, 

2x« + X + 3 = -4x + 5+ (Cx + D)(x2 + 1). 
Equating the coefficients of like powers of x and solving, we get 

^=-1, B = 3, C = 2, D=0. 

Hence r2x»+^+8 ^ f-^ + S /• 2xdx 

•^ (x2 + 1)2 J (x» + 1)« J xa + 1 

= log(xN-l)+/^,dx. 

Now apply formula (J?), p. 326, to the remaining integral. Here 

-4 = - 1, 5 = 3, i) = 0, g = 1, n = 2. 
Substituting, we get 

/•-X4-3, l + SxSfdx 1 + 3x3 

I — -ox = + - I = + - arc tan x. 

J (x2 + 1)4 2(x2 + 1) ^ 2 J x2 + 1 2(x2 + 1) 2 

Therefore 

f2x8 + x + 3 I , o . ^x l + 3x 3 

I — rs — TTTT- = log(xa + 1) + — -T + - arc tanx + C. 

J (x2+l)a '^^ ^ ' 2(x2 + l) 2 



1. r?l±£ — }_^ _ _? — ?_ _^ iQg /^.a ^ 2)* — arc tan — + C. 

J (x2 + 2)a 4(x2 + 2)^ ^^ ^ ' 4v^ V^ 

2. r — ^^^ =i,og^L±i+_^ij_+c. 

J (1 + x){l + x2)a 4 ^ (X + 1)2 2{x» + 1) 

rx7 + x» + x« + x ^ 6 _10_ l»j^^ 2 2) - 91og(x2 + 3) + C. 

J (xH2)2{xa+3)2 2(xa + 2) x3 + 3 2 *^ ^' 6V t/t 

. f (4x2-8x)dx 3x2-x . (x-l)a . ^ .^ 

4. I — ^ = hlog^^ ^ + arctanx + C. 

J (X- l)a(x2 + 1)2 (X - 1) {x2 + 1) ^ x2 + 1 

- r (Sx-\-2)dx 13X-24 .26 ^ 2x-3 . _ 

6. I — ^^ = J- arc tan \- C. 

J {x2 - 3x + 3)2 8(x2 - 3x + 3) 3 V3 V3 

Since a rational function may always be reduced to the quotient 
of two integral rational functions (§ 26, p. 16), i.e. to a rational 
fraction, it follows from the preceding sections in this chapter 
that any rational function whose denominator can be broken up 



828 INTEGRAL CALCULUS 

into real quadratic and linear factors may be expressed as the 
algebraic sum of integral rational functions and partial fractions. 
The terms of this sum have forms all of which we have shown 
how to integrate. Hence the 

Theorem. The integral of every rational function whoBe denomi- 
nator can he broken up into real quadratic and linear factor% may 
he founds and is expressible in terms of algebraic, logarithmic, and 
inverse-trigonometric functions; that is^ in terms of the elementary 
functions. 



CHAPTER XXVII 

INTEGRATION BY SUBSTITUTION OF A NEW VARIABLE. 

RATIONALIZATION 

195. Introduction. In the last chapter it was shown that all 
rational functions whose denominators can be broken up into real 
quadratic and linear factors may be integrated. Of algebraic 
functions which are not rational^ that is, such as contain radicals, 
only a small number, relatively speaking, can be integrated in terms 
of elementary functions. By substituting a new variable, however, 
these functions can in some cases be transformed into equivalent 
functions that are either in the list of standard forms (pp. 292, 293) 
or else are rational. The method of integrating a function that is 
not rational by substituting for the old variable such a function of 
a new variable that the result is a rational function is sometimes 
called integration by rationalization. This is a very important arti- 
fice in integration and we will now take up some of the more 
important cases coming under this head. 

196. Differentials containing fractional powers of x only. 

Siich an expression can be transformed into a rational form by 

means of the stibstitution 

a: = e", 

where n is the least common denominator of the fractional exponents ofx. 
For x, dx^ and each radical can then be expressed rationally in 
terms of .. 

—dx. 

Solution. Since 12 is the L.C.M. of the denominators of the fractional expo- 
nents, we assume x — z^ 

Here dx = 12 z^kte, x' = z^, x* = 2», x* = sfi. 

.-. r ^ ~^ dz = r ^~^' 12 2"dz = 12 Hzw - z8) dz 

= f z" - I z« + C = f X* - 1 X* + C. 

fSabetitating back the value of z in terms of x, namely, z=x'^.l 

329 



830 INTEGRAL CALCULUS 

The general form of the irrational expression here treated is then 

I 

where R denotes a rational function of o^. 

197. Differentials containing fractional powers of a + to only. 

Such an expression can be tran^ormed into a rational form by 

means of the substitution 

a-{-bx = z\ 

where n is the least common denominator of the fractional' exponents 
of the eapression a -\- bx. 

For Xy dxj and each radical can then be expressed rationally in 
terms of .. 

Ex. 1. Find ^ ^ 



r ax 

•^(H-x)* + (l + x)*' 



Solution. Assume 1 + x = 2^ ; 

then dx = 2 zdz, (1 + x)' = ««, and (1 + «)* = «. 

dx r 2zdz _ r dz 

(1 + a;)i -f (1 + X)* ""'^ «• + « ~ J z^-hl 

= 2 arc tan 2 + C = 2 arc tan (1 + x)* + C, 
when we substitute back the value of z in terms of x. 

The general integral treated here has then the form 

I 

-B [a:, (a -h bxy] dx^ 
where B denotes a rational function. 



'}dz 4 






;*-x* , 1/2 . 6 
6x^ 
;* + 1 ^ 6 12 



3. r^Jl-dx = --- + — + 21ogx-241og(xA + i) + C. 

•^ X* + X* X* X" 

/elf ft f^ 1 

— = -X* + 2 log "" + 4 arc tan x* + C. 
X" — X* «* X* + 1 



SVxdx ,«rx' x' 4x* 



/avxox ,„rx* x« 4x' 11 1 T 



INTEGRATION BY RATIONALIZATION 831 

•^aj{x + l)* (x + l)*H-l 

"•^(o + te)* ftaVa + to ' '•^(4x + l)* 12(4x + l)* 

0. Cy Vo H- y dy = A (4 y - 8a) (a + y)* + C. 

1. r ,^jti'^^ dx = x + l + 4Vx + l + 41og(Vx + l^l) + C. 
•^ Vx + 1 - 1 

2. f ^= = ?{x + 1)« - 3(x + 1)* + 31og(l +^^iTl) + C. 

•^ 1+Vx + l 2 

. f ^ 1 = 8 {(X + 1)* + 2 (x + 1)» + 2 log [(X + 1)* - 1]} + C. 

•^ (X + 1)' - (X + 1)* 

L98. Differentials containing no radical except Va + bx + acK* 
Such an eapression can be transformed into a rational form by 
means of the substitution 

Va -\'bx-\-3[? = 2 — a:. 

For, squaring and solving for x, 

^ = ^; thenefe=2(zl±^^ 

and Va4-6x + a:'(=g~a?)= , J" — 

^ b-\'2z 

Hence x^ dx^ and Va -\-bx-\-:3^ are rational when expressed in 
terms of z. 

-== 

V 1 + X + X* 

Solution. Assume Vl + x + x* = « — x. 

Squaring and solving for x, 

2z + l {22 + 1)-^ 

and vT+x + x2(==2--x)== V^— ^. 

^ ' 2« + l 

• If the nullcal is of the form Vn +px + qx*^ (7>0, it may be written Vgx/'* + ^x + x«, and 

therefore oomes under the above head, where a = - 1 6 = - • 

9 9 



332 INTEGRAL CALCULUS 

jf::!>; =/-f^=iogc(2,.n)c] 

2« + l ^ 

= log[(2x + 1 + 2 Vi + a; + x«)c], 

when we substitute back the value of z in terms of x. 



199. Differentials containing no radical except Va + 6x — «■.* 
aSWcA aw expression can be transformed into a rational form by 
means of the substitution 

•\/aJ^bx-a? [= V(a: - a) (/S - a:)] = (x - a) 2, 

wAere a; — a an<i /8 — a: are real^ factors of a-\-bx — a^. 

For, if -Va-^-bx — a? = V(a: — a) (/8 — a;) = (a? — a)^, 
by squaring, canceling out {x — a), and solving for a;, we get 

a: = ^;theni. = ^(;^-y , 

and VaHri^^^[= (a: — a)2j] = ^ ,"" V^ * 

Hence «, rfa;, and Vo-f-fti^-^ are rational when expressed in 
terms of 2. 

dx 



/ax 
V2 + X - x* 



^Solution. Since 2 4- « - a:* = (x + 1) (2 — x), we assume 

V(x + 1) (2 - X) = (X + 1)2. 

2 - 2« 
Squaring and solving for x, x = — — - . 

2 "T" A 

-62d2 __../:;—-: — -Sr . . .. , 3« 



— Q ZtiZ / 

Hence dx = — - — — » and v2 + x - x« [= (x + 1)2] = 



(22 + 1)3 '■''"' 2« + l 

/dx ^ r dz 

-;== = - 2 I -— - =- 2 arc tan2 + C 
V2 + X - xa ^ 2« + 1 



= — 2 arc tan \/ + C, 

when we substitute back the value of z in terms of x. 



/2~x 



• If the radical is of the form Vn+px - qjc», g >0, it may be written V? \/- + - x - «•, and 

therefore oomm under the above head, where a=-ib=-- 

g 9_ 

t If the factors ota-i-bx-x* are imaginary, Va + 6x-ar* is imaginary for all ralaes of x. For, 
if one of the factors is x - m + in, the other must be - (x - m - in)^ and therefore 

6 + ox -x* = - (x-m-¥- in) (x-m-in) — - [(x—m)* + n*], 

which is negatiye for all values of x. We shall consider only those cases where the factors are reaL 



INTEGRATION BY RATIONALIZATION 833 

EXAMPLES 

Vxa - X + 2 +a:-V2 



, /• dx 1 , Vxa - X-I-2+X-V2 ^ _ 

1. ) — ; = "-F tog -7: 7: + C. 

•^ X V x* - X + 2 V 2 V x2 - X + 2 + X + V 2» 



dx 



2. f — ^ = 2arc tau(x + Vx2 + 2x-l) + C, 
^ xVxaH-2x-l 

3. r__^^^ = _2arctanfLz^y + C. 
•^ V2 - X - x2 \x + 2/ 



dx 



4. f ^ = log(2 Vx5« - X - 1 + 2x - 1) + C. 

•^ Vx2 - X - 1 

* r <^ « /l — as ^ 

6. I -—== = - 2arc tan -v/ ^ + C 

^ V4x-3-xa \x-3 

g r xdx ^ 8-K6X ^^ 

* -^ (2 + 3x-2a^)* 26V2 + 3x-2x» 

7. r ^<^ = V6 log(6x - 1 + V6 V6xa-2x + 7) + C. 

^ V6xa-2x+7 

8. f ^ — = -i,log(6x - 1 + 2 V3 V8x2-xHH) + C. 
•^ V3x»-x + l V3 

9. f-— i^^=2%^arc8infi^) + C. 
•^ V4H-3x-2x^ . ^ V4I ^ 

10. r_^=, = log(l + x + ViM^) + C. 

11. r<lfL±^)^ = log(x + l+V2^T7a) i -fC. 

•^ x^ X + V2x + xa 

The general integral treated in the last two sections has then 

where R denotes a rational function. 

Combining the results of this chapter with the general theorem 
on page 328, we can then state the following 

Theorem. Every rational function of x and the square root of a 
polynomial of degree not higher than the second can be integrated 
and the result expressed in terms of the elementary functions.* 

* As before, however, it is assumed that in each case the denominator of the rational function 
can be broken up into real quadratic and linear factors. 



334 INTEGRAL CALCULUS 

200. Binomial differentials. A differential of the form 

where a and b are %ny constants and the exponents m, n, je> are 
rational numbers, is called a binomial differential. 

Let a: = z*; then dx = a«*"*(fe, 

and or (a 4- bif'ydx = a«-^+— ^(a + bsT^ydz. 

If an integer a be chosen such that ma and na are also integers,* 
we see that the given differential is equivalent to another of £he 
same form where m and n have been replaced by integers. Also, 

or (a -h b3fydx = oT'^'^iax-'' + bydz 

transforms the given differential into another of the same form 
where the exponent n oix has been replaced by — n. Therefore, 
no matter what the algebraic sign of n may be, in one of the two 
differentials the exponent of x inside the parenthesis will surely 
be positive. 

When |> is an integer the binomial may be expanded and the 
differential integrated termwise. In what follows p is regarded 

as a fraction; hence we replace it by -, where r and % are integers.! 

We may then make the following statement : 

Every binomial differential may be reduced to the form 

r 

of (a -{- bif)' dxj 
where w, w, r, 8 are integers and n is positive, 

201. Conditions of integrability of the binomial differential 

r 

(A) «"* {a + 6x~)* dx. 

Case I. Assume a-^-baf = 2^, 

1 r 

Then (a + baf)' = 2, and (a -f bsf)' = s^; 

I m 

also x=( T" j , and af* = ( ~ j ; 

hence dx = -r-^'^ ( 7 1 ^^• 

071 \ b J 

* It is always possible to choose a so that ma and na are integers, for we can take « as the 
L.C.M. of the denominators of m and n. 

t The case where p is an integer is not excluded, but appears as a special case, yis., r»p, • « t. 



I a 



INTEGRATION BY RATIONALIZATION 835 

Substituting in (A)^ we get 



The second member of this expression is rational when 

Tn + l 

n 
is an integer or zero. 

Case IT. Assume a-^-baf" = ^'af. 

Then af = r » and a -h 62f = ^^af = 



r r r 

Hence (a + bo*)' = a'(z^- b) V; 

1 1 M m 

also a; = a»(2'-6)"% a:" = a"(2-- J)"»; 

« 1 -1-1 

and (fo = --a"2;-^(2;'-J) " rfa?. 

n 



Substituting in (A)^ we get 



8 —r- + z 



+ I . *• /ne-\- 1 . r 






a:-(a + 6af)'rfa; = --a " •(z'-J) v « . ^z--*-'-^dz. 

n 

The second member of this expression is rational when — — — \- - 
. . n 8 

IS an mteger or zero. 

Hence the binomial differential 

a:^ (a -f b^ydx 

can be integrated by rationalization in the following ca8e8:* 

Case I. When — ^^^— = an integer or zero^ by a88uming 

n 

a -{- baf* =^ z^. 

m -I- 1 ?* 
Case II. When — — — \.- = an integer or zero^ by a88uming 

n 8 

a -f bsf = z*a:*. 

* Annmlng as before thAt the denominator of the resulting rational function can be broken 
up into real quadratic and linear factors. 



L 



886 INTEGRAL CALCULUS 

EXAMPLES 

•^ (a + tea)* ^ ^ Vo + tea 

So2u^ion. m = 8, n = 2, r= — 8, 8 = 2; and here = 2, an integer. 

Hence this comes under Case I and we assume *'' 

— - — ) , dx = -; -1 and (a + te*)" = «•. 

/ x^dz _ r/ z^ — a \* «2z 1 

(a + te«)*~*^^ * ^ 6*(za-a)* «• 

1 2a + 6x2 



^ Va + te» 



+ C. 



" ^x* 



dx ^ (2xa-l)(l+x«)* 
VT+^'" 3x» 



+ C. 



Solution. m = -4, n = 2, - = — j and here 1- - = — 2, an integer. 

8 2 n s 

Hence this comes under Case II and we assume 

l + x« = «2x2, « = (L±^. 

X ' 

whence «■ = -:; — r» l + x« = -- » Vl + x^ = 



2a_i ^ 2a_i ^ («'-l)* 



1 1 , afa 

also X = , X* = — ; and dx = — 



zdz 
r dx I (2*"1)' /• 



(Z«-l)« (,S_1)1 

_ z» , ^_ (2x'-l)(l + x«)* , „ 



•^ 16 

4. C^L- = ..^^=^ + c. 6. r-^ = (i + «.)i<^ + c. 

•^ (1 + x2)' Vl+x2 -^ Vl+xa 3 

^ r dx 1, ex 

6. j — -p=:^=i = -l0g- 

•^ X V a2 - x« « Va2 - x^ + a 
7. r__^^ = -a(l+xT*(2x + l) + C. 



INTEGRATION BY RATIONALIZATION, 887 






9. rt»(l + 2«2)^(tt = (l + 2«»)«^^- — ?:+C. 

10. ru(l + w)*du= A(H-u)*(5u-2) + C. 
a^da a* 



11. f-^ 



+ C. 



6a2)* 3a(a + 6o2)* 
12. /(^ (1 + <^' (W = 5»j (1 + <^) V - i (1 + ^)» + A (1 + ^* + C. 



dx _ 3g» + 2a 



13. r ^ , = "-^^^^^ -fC. 



202. Transformation of trigonometric differentials. 

From Trigonometry 

(A) sin a: = 2 sin -cos-, 87, p. 2 

(B) cos X = cos* - — sin* - . 87, p. 2 

a? 

jy , , X 1 1 2 

But sm - = 

2 x 



•^^2 V^^'l-^' >/l+*^ 



X 1 

and cos - = - 



2 



If we now assume 



-w 



sec^ 'l+tan*| 



tan ^ = 2? or, x = 2 arc tan 2, 

X Z X A. 

we get sin - = , , cos ^ = , 

Substituting in (A) and (5), 

2^ 1-2* 

sma: = r ;, cosa; = 3 ;• 

2d2 

Also by differentiating x=2 arc tan z we have dfx = 3 ^ 



388 ^ INTEGRAL CALCULUS 

Since sin Xj cos s^ and dx are here expressed rationally in terms 
of 2, it follows that 

A trigonometric differential involving sinx and cosx rationally 
only can be trantformed by means of the mbstitution 

tan- = j5, 

orj what is the same thing^ by the svhstitutions 

2z 1-z^ J 2dz 
sm2:=:3 :;> cosa: = r r, aa; = 



1+2* 1+2" 1+2" 

into another differential expression which is rationed in z. 

It is evident that if a trigonometric differential involves tan z, 
cot Xj sec x^ CSC x rationally only, it will be included in the above 
theorem, since these four functions can be expressed rationally in 
terms of sin Xj or cos x^ or both. It follows therefore that any 
rational trigonometric differential can be integrated.* 



EXAMPLES 



I. I ^ ^ ^ =-tan«- + tan-4--logtan^ + C. 
J sin X (1 + C08 sc) 4 2 22^ 2 



(1 + COB x) 

SoliUum, Since this differential is rational in sin x and cos x, we make the ahore 
substitutions at once, giving 



/ (l + 8inx)dx _ / \ l + z'/14-a 
sinx(l + co8x) " I 22 / 1 -gg 

•^ l + zA "'"l + ga 



) 

1 *X X 1 . / x\ ^ 



tan? -2 



2. r_^=!iog_J— +0. 

•/4 — 6smx 3 1^. X .1 



2tan--l 
2 



• See footnote, p. 330. 



INTEGRATION BY RATIONALIZATION 889 



3. f = -arc tan (^2 tan -Wc. 

J 6-3C08X 2 \ 2/ 

. /• dx 1 ^ /6tanx + 4\ , ^ 

4. I -——-— = - arc tan ( ) + C. 



dx 1 



5 C = - arc tan(3 tan x) + C. 

J6-4co62x 8 ^ ' 



tan- + 2 

6. (- — = 7log +C. 

^3 + 6coflX 4 \ X „ 

tan — 2 

2 



/Binxdx 2 , _ * /* asx , ^ 
= _^_ + 2 arc tan ( tan - J + C 
l+sinx . . , X \ 2/ 
1 + tan- 



^ /• C0BX(2x ^ / x\ X « 

a I = 2arctan(tan-)-tan- + C. 

Jl + cosx V 2/ 2 

9. Derive by the method of this article formulas [16] and [17], p. 208. 

203. Miscellaneous substitutions. So far the substitutions con- 
sidered have rationalized the given differential expression. In a 
great number of cases, however, integrations may be effected by 
means of substitutions which do not rationalize the given differen- 
tial, but no general rule can be given, and the experience gained 
in working out a large number of problems must be our guide. 

A very useful substitution is 

1 , dz 

Z 2* 

called the reciprocal substitution. Let us use this substitution in 
the next example. 



Vaa-x» 

-ox. 



Ex. 1. Find /— ^ 

X* 

I dz 

Solution. Making the substitution x = -f dx = — -,weget 

z z* 

J X* */ ^ ' SaJ 3aax» 



840 INTEGRAL CALCULUS 



1. f — — = — log +C. AaBamex' = £. 

Jx(a» + x») 3a« *o« + x« 






AflBume X — 2 = z. 






+ log(x + 1) + C. Assume x + 1 = £• 



6{x + l)8 
— — — = ^ + C A88nmex=:- 



_ /• dx 1, ex 

6. I , = -log 

•^ X Va« + x* <» a + V o2 + x« 



Assumo X = — • 



6. f-^^ = |(x«-8)(x« + l)* + C. Assume x* + 1 = «. 

•^ (x^ + 1)* 8 

/(2x ex 1 

— - = log - » Assume x = -' 

X Vl + X + x2 2 + x + 2vl+x + x^. « 

8. /^-ti?i^dx = ?(l + logx)* + C. Assume 1 + log x = «. 

X u 

= --(3c* - 4) (c + 1)« + C. Assume c* + 1 = «. 

(e* + l)* 21 

1^' f-i ;r- = :;^ ^ + -log(e«-2) + C. Assumee« = «. 

J ^x_2e* 2e* 4 4 ' 

/x'^dx 1 X Vl jC* 
, = - arc sin X h C Assume x = cos z. 

Vl-x« 2 2 

va* — x"dx = — arc sin - -(- - va'-' _ x^ + C. Assume x = a sin t. 

2 a 2 



CHAPTER XXVIII 

INTEGRATION B7 PARTS. REDUCTION FORMULAS 

2S01 Formula for integration by parts. If u and v are functions 
of a single independent variable, we have from the formula for the 
differentiation of a product, V, p. 144, 

d{uv) = lulv + vduj 
or, transposing, 

udv = d (uv) — vdu. 

Integrating this, we get the inverse formula, 
{A) I udv = ut; — I vdUf 

called the formula for integration by parts. This formula makes 
the integpration of udvj which we may not be able to integrate 
directly, depend on the integration of dv and vdu^ which may be 
in such form as to be readily integrable. This method of integror 
Hon hy parts is one of the most useful in the Integral Calculus. 

To apply this formula in any given case the given differential 
must be separated into two factors, namely, u and dv. No general 
directions can be given for choosing these factors, except that 

(a) dx is always a part of dv ; and 

(6) it must he possible to integrate dv. 

The following examples will show in detail how the formula is 
applied. 

Ex. 1. Find jxcoBxdx. 

Sclviion. Let u = x and do = cos idx ; 

then du = dx and v = fcos xdx = sin x. 

Sal)Btitating in {A), 

u dv u V V du 

z C08 xdx = X sin x — I sin x dx 

= X sin X + cos X + C. 
341 



332 INTEGRAL CALCULUS 

/ 2(z« + z + l)dg 

= log[(2x + 1 + 2 Vl + x + x»)c], 
when we substitute back the value of z in terms of x. 



199. Differentials containing no radical except Va + bx — as*.* 
aS'uc^A aw expreBsion can he transformed into a rational form by 
means of the suistittUion 

Va-l-62: — a:*[=V(ic -~a){^ — x)] = (x — a)«, 

wAer« a; — a a?ui )8 — a; are real ^ factors of a + hx — of. 

For, if Va -f Jar — ar* = V(a; -^a){fi^ x) = (2; — a)2, 
by squaring, canceling out (a; — a), and solving for x, we get 

and Va -f 6a: - a:^r= (a? - a)2] =i^fl^. 

Hence x^ dx^ and Va -{-bx^a? are rational when expressed in 
terms of 2. 

/dx 
-7 
V 2 + X - x^ 

(SoZti^ion. Since 2 + x — x^ = (x + 1) (2 — x), we assume 

V (X + 1) (2 - X) = (X + 1)2. 
2-«2 



Squaring and solving for x, x = 



z^+l 



Hence dx = /^^f , and V2 + x-x«[= (x + 1)«] = ^' 



. = - 2 I -«—T =-2arctaji2 + C 

V2 + X - xa -^ 2^ + 1 , 

12 —X 

= — 2 arc tan \ H C, 

\x + 1 
when we substitute back the value of z in terois of x. 



• If the radical is of the form Vn+px - qx^, 9 >0, it may be written y/q \/^ + - x - «•, and 
therefore oomea under the above head, where a= - f 6= - • 

q q__ 

t If the factors of a + 6ar - ar* are Imaginary, Va+6a:-jr» is imaginary for all values of x. For, 
if one of the factors is x - m + in, the other roust be - (x - m - xn)y and therefore 

6 + ax-x* = -(x-m + \n^ (x-m- in) = - [(x - m)* + n*], 

which is negative for all values of x. We shall consider only those cases where the factors are real. 



INTEGRATION BY RATIONALIZATION 333 



1. I —. = = -7=^<^g" 7 F+^- 

•^ X Vx* - X + 2 v2 V x2 _x-f2 + x + v2» 

2. f — ■ = 2arc tau(x 4- Vx^ 4- 2 x - 1) 4- Cf. 
•^ xVx2 + 2x-l 

3. f ^ =-2arctan(i— ^) + C. 
•^ V2 - X - x^ \x + 2/ 

4. f , ^ - = log{2 Vx'* - X - 1 4- 2x - 1) + a 
•^ Vx« - X - 1 

5. r , ^ — =~-2arctan^/l^-l-rf. 
•^ V4x~3-xa \x-3 

/• xdx 8 4- 6x ^ 

^ (2 + 3x-2x2)i 26V2 + 3x-2x« 

7. f , ^^ =V51og(fix-l^V5Vfix2-2x4.7)^rf. 

^ V6xa-2x+7 

•^ V3x«-x + l v3 

9. f A^__=2V2arcBin(i^Uc. 
•^ V4 + 3x-2x2 . ^ Vil ^^ 

10. f -^^=^ = logf^ + X + Vx» + x) + C. 
•^ Vx« + x ^2 / 

11. r(lfL±^)i^ = iog(x + i+V27T7«) i + c. 

-^ xa x + V2x-f xa 

TThe general integral treated in the last two sections has then 
the form ^ ^^ -sla^hx-^c^) dx, 

where R denotes a rational function. 

Combining the results of this chapter with the general theorem 
on page 328, we can then state the following 

Theorem. Every rational function of x and the square root of a 
polynomial of degree not higher than the second can be integrated 
and the result espressed in terms of the elementary functions,* 

• Ag before, however, it is asanmed that in each caae the denominator of the rational function 
can be broken up into real quadratic and linear factors. 



344 INTEGRAL CALCULUS 

13. fx.e«d, = !^Cx.-?^ + »f-A) + c. 
J a \ a a> a'/ 

14. r 0> sin 0d0 =:2co8 + 20Bin0-0Sco8 + C. 

16. r(logx)a(to = x[loga«-21ogx + 2]4-C. 

/ft* 
a tan* ada = a tan o — — + log cos ft + C 

"■/^^■-.-fT'°" -■"«'- ■> + <'• 

dx 
Bint. Let u=« log x and d» =• 1 etc. 

/x' X* + 2 / 

x«arc8inx(ix = --arc8inx4- — ^— vl-x«+ C. 
3 

19. r8ec9 0logtan9d« = tantf(logtan^- 1) + C. 

/dx 
log (log x) — = log X • log (log x) — log X + C. 
X 

21. rM+il^ = 2 v5rTi[iog(x+i)- 2] + c. 

•^ Vx + 1 
22. fx^{a - x«)*dx = ~ J x2(a - x«)« - ,J^ (a - x2)i + C. 

lRn<. Let u => a;> and dv » (a - a;*)' xdXt eto. 

•^ x' Sx^L ^ ^J 

205. Reduction formulas for binomial differentials. It was shown 
in § 200, p. 334, that any binomial differential may be reduced to 

where p is & rational number, m and n are integers, and n is posi- 
tive. Also in § 201, p. 334, we learned how to integrate such a 
differential expression, in certain cases. 

In general we can integrate such an expression by parts, using 
(-4), p. 341, if it can be integrated at all. To apply the method 
of integration by parts to every example, however, is rather a long 
and tedious process. When the binomial differential cannot be 
integrated readily by any of the methods shown so far, it is cus- 
tomary to .employ reduction formulas deduced by the method of 
integration by parts. By means of these reduction formulas the 
given differential is expressed as the sum of two terms, one of 



REDUCTION FORMULAS 345 

which is not affected by the sign of integration, and the other is 
an integral of the same form as the original expression, but one 
which is easier to integrate. The following are the four principal 
reduction formulas. 

[A] Cx^ {a + hx*^)P dx 

[E] Cx"^ (a + bx^)Pdx 

= ^ — ' — f- ^- I a^ (a + hx^\P'-^dx% 

np + t» + 1 np + «» + 1 ^ 

[C] Cx^(a + bx^)Pdx 

(w + l)a (w + l)a J 

[I>] fa?"* (a + hx'^)P dx 

n(jp + l)a n(p + l)a J 

While it is not desirable for the student to memorize these 
formulas, he should know what each one will do, namely: 
Formula [A] diminishes mhy n. 
Formula [J5] diminishes p hy 1, 
Formula [C] increases m hy n. 
Formula [I>] increases p by 1. 

I. To derive formula [A], 

The formula for integration by parts is 

(A) j udv = uv-' i vdu. (-4), p. 341 

We may apply this formula in the integration of 

CQir{a-\-h7rYdx 

by placing u = a:^""^^* and dt; = (a 4- bafyaf^dx; 

then (iw = (7n-n + l)rr- ia: and v = ^?-t_^!)^. 

nb{p-^l) 

* In order to integrate dv by [4] it is necessary that x outside the parenthesis shall have the 
exponent n -1. Subtracting n - 1 from m leaves m-n + l for the exponent of x in u. 



846 INTEGRAL CALCULUS 

Substituting in (A)^ 
(B) Car {a -{- baf'y dx 

wi(j? + l) nb{p + l)J ^ ^ 

But (V -" (a -f 6x-)'» +^ (ia; =f^~* (« + ia:")" (« + ia:") dx 

= a Ta:*— (a -f 6af')''cfo + h]Qr{a + harydx. 
Substituting this in (5), we get 

Transposing the last term to the first member, combining, and 
solving for |V'(a 4- haf^ydx^ we obtain 

[A\ fx^ {a + bx^)P dx 

b(np + m + l) b{np + m + l)J k t ^^ } 

It is seen by formula [A] that the integration of af"(a 4- la^ydx is 
made to depend upon the integration of another differential of the 
same form in which m is replaced by m — n. By repeated appli- 
cations of formula [-4], m may be diminished by any multiple of n. 

When rip 4- w -h 1 = 0, formula [-4] evidently fails (the denomi- 
nator vanishing). But in that case 

m4-l ^ 

n 

hence we can apply the method of § 201, p. 336, and the formula 
is not needed. 

IL To derive formula [B], Separating the factors, we may write 
((7) (3r{a 4- ha^ydx =f^{a 4- baf^y-^ (a 4. baf) dx 

= a Car (a 4. bafy^dx 4. bCar+''{a 4- 6a:")''-^ia;. 



REDUCTION FORMULAS 847 

Now let us apply formula [A] to the last term of (C) by substi- 
tuting in the formula m + n for m and j? — 1 for p. This gives 

Substituting this in ( (7), and combining like terms, we get 

[B] Cx^ (a + 6a5*») p da> 

= ^ — • ^—4- ^- \ x^(a 4-bx^)P''^dx9 

np + fn + 1 np + fn + IJ 

Each application of formula [B] diminishes j? by unity. Formula 
[B] fails for the same case as [A], 

III. To derive formula [C]. Solving formula [A] for 

CoT'* {a + bsfydx, 
and substituting m + n for m, we get 

[C] ffic*** (a + bx^)Pdx 

a(m + l) a(m + l) J \ t / 

Therefore each time we apply [C], m is replaced by m -f n. When 
m-f 1 = 0, formula [C] fails, but then the differential expression 
can be integrated by the method of § 201, p. 885, and the formula 
is not needed. 

IV. To derive formula [!>]. Solving formula [B] for 

and substituting p -\-l for j?, we get 

[I>] Cx*^ {a + bx^) P dx 
^_xn^^Ha + bxn)P^^ np + n + m + 1 r^n.^a+bxn)P^^dx. 

Each application of [I>] increases p by unity. Evidently [I>] 
fails when |? + 1 = 0, but then j? = — 1 and the expression is 
rational. 



348 INTEGRAL CALCULUS 



1. r_^^ = _l(je24.2){l-x«)* + C. 
J Vi — X* ° 

SohAifm, Here m = 3, n = 2, p = — J, o = 1, 6 = — 1. 

We apply reduction formula [^i] in this case because the integration of the 
differential would then depend on the integration of \v,{y — x>)~^(2x, which comes 
under [4], p. 292. Hence, substituting in [^i], we obtain 

f-^/t •v-*^ aj»-2+i(l -aJ^)"*"*"^ 1(3-2 + 1) f^ ,,, ^v-4^ 

= _ 1x2(1 - x»)* 4- f /«{! - xT**B 

= - }x»(l - x«)* - i(l - x«)* + c 
= -J(«* + 2)(l-x2)* + C. 

2. f-^!^— =-flx8 + ?a»x)V^"^^ + |a*arc8in? + C. 
•^(a2-x«)* V4 8 / 8 a 

Exat, Apply [il] twice. 

3. r(a« + x«)*dx = ? Va» + x* + ~ log(x + Va* +x2) + C. 

•/ A 2 

^n<. Here m= 0, n= 2,ps }, a= a*, &« 1. Apply \B\ onoe. 

. /• dx (X2-1)* 1 ^^ 

4. I . = ^ -r- + - arc sec X + C 

^ x^yf^^^ 2x3 2 

Hint, Apply [C] once. 

5. I , = V a^ — X* H — arc sm - + C7. 

•^ Va« - x» 2 2a. 

6. f ^!^ = l(x«-2a2)Vag-f x« + C. 
•^ voa + xa 3 

•^ Vr^ \ 6 16 15/ 

8. fx^ Va»-x2dx = ^ (2x2 - a«) Va^ - x» + -arcsm ? + C. 
•/ 8 8 a 

mnr. Apply [i<] and then [^. 

9. I = h — arc tan - + C. 

J (a* + xa)a 2a2(oa + x^) 2a» o 

JERfi^ Apply [/)] onoe. 

/■ dx _ y/ofi^^^ 1 X 

' Jx'VS^rr^a" 2a'^2 ■^2a» ""^a + V^T^a"^^* 



REDUCTION FORMULAS 849 

12. r ^ =(-^^'-^^')%c. 

13. r(x« + a2)»da; = 3(2x» + 6a«) Vx^ + a« + ^log(x + Vx^ + a«) + C. 

14. rxa(x« + a^^dx = ?(2x» + a«) Vx^ + a^ - ^ log(x + Vx^ + oS) + C. 
•/ 8 8 

16. I = -^- — {2ax-aJ«)* + — arc vers- + C. 

JWnl. f— ^^=:- fxl(2a-«)-4AF. Apply [^ twice. 

/dz Va* — X* 

x»(o«-x«)* o** 

•^ V2fy-y« ® 2 r 

18. f ^ ^ = - (2 at - «*)* + oarc vers - + C. 

19. I = 1 arc tan - 4- C 

J (a2 + a2)8 4o»(aa + ««)» 8a*(as + a'^) 8a» a 

20. f ^**^^ = - A (3 ,« + 4r8 + 8) Vl - r« + C. 



206. Reduction formulas for trigonometric differentials. The 

method of the last section, which makes the given integral depend 
on another integral of the same form, is called 9ucce89ive redv/ition. 
We shall now apply the same method to trigonometric differentials 
by deriving and illustrating the use of the following trigonometric 
reduction formulas. 



8in*~x cos** X dx 

8in*~ + *a5C08»*-^a5 . n— 1 



+ 5: r 8in*»*a5C08»*-*a5<to. 

m + n tn + nj 

+ ^^^^ — I 8in*»-* 0? QOS~ X (fa?, 
fn + n m + nJ 



sin*** X cos* X dx 

8in'"-^ficco8**+^a5 . f»— 1 



860 INTEGRAL CALCULUS 

[ G] j 8in** X C508* X dx 

= + — -J 7 — I 8m**a5C508*+*x«a5« 

n + 1 n + 1 J 

[ jET] j 8ln'* X C08" X dx 

^ 8lii"*^ixco»-^xa, w + n + a r8to«.+.a,co»«a5*D. 

W»+ 1 t» + 1 J 

Here the student should note that 

Formula [E] diminiihes n hy 2. 
Formula [-F] diminishes m by 2. 
Formula [O] increases n by 2. 
Formula [H] increases m by 2. 

To derive these we apply as before the formula for integration 
by parts, namely, 

(A) j udv =zuv— i vdu. (^), p. 841 

Let ii = cos"~*a:, and <iv = sin*" 2; cos a;dar; 

then du = — (n — l)cos''"*a;8ina^a::, and v = 



sm"''^*2; 



m + 1 

Substituting in {A\ we get 

(B) I sin"* a: C08"a^ = -f 



8in""*"*a;co8"~'a; 



H :r 1 sin'"'*''a;cos""*a:diB. 

m4- 1*^ 



7n-f 1 

n-^1 
m-{- 
In the same way, if we 

let u = sin'*~^2^ and £2t; = co8"2:sinx(2z^ 

we obtain 

/•vx r • « -J sin'""^iccos''"*"*a; 

( C) I sm"'ic cos* a:aic = z 

^ •/ n 4- 1 

m — 1 
n + 



-f 3- I 8in"'"'a:cos"'*"*a;diB. 

n4- !•/ 



But J 8in'""*'*a; C08""*a;da; = I sin"*a:(l — cos^x) C08*"'xia? 

= j sin* a: cos"~* xdx — I sin"'^: cos*xdx. 



8in*^a; cos*^a? duo 

8in*"+^a5co8*-*a5 . n— 1 



REDUCTION FORMULAS 361 

Substituting this in (5), combining like terms, and solving for 
J sin* a: cos'a^rfa:::, we get 

, H ^T — I 8ln*»a5C08*-*a5cla5. 

tn + n tn-^nj 

Making a similar substitution in ((7), we get 

[F] I sin** as cos* a? ciai; 

8in*»*-*a5co8*+^a5 ,«» — !/•*«»• « ^ 

tn + n tn -^nj 

Solving formula [E] for the integral on the right-hand side, and 
increasing n by 2, we get 

[O] I sin** X cos** X dx 

»in**+*a5COS**+^x , m + n 4-2 r * ^ «...• ^ 

= rri . 7^ — I sin*** 35 cos*»+* a? <la?« 

n+1 n+1 J 

In the same way we get from formula [F]^ 

[H] I 8in*'*a5 cos** a? dx 

8in***+*a?cos**+^i» «» + n + 2 /• . ^^, ^ , 

= r-z . J — I sin***+«a5co8**a5<fa?. 

«» + 1 «» + 1 J 

Formulas [-B] and [J'J fail when m 4- n = 0, formula [O] when 
n 4- 1 = 0, and formula [H] when m 4- 1 = 0. But in such cases we 
may integrate by methods which have been previously explained. 

It is clear that when m and n are integers the integral 

J sin"'a:cos''a:di 

may be made to depend, by using one of the above reduction 
formulas, upon one of the following integrals: 

dx r dx 



/dXf iQinxdxj icoQxdx^ I sin a; cos zdo:, I -: » 1- 
J J J J sin X J ( 



sin X •/ cos X 

dx 
cos X sin X 

all of which we have learned how to integrate. 



: — » I tan xdx^ \ cot xdx^ 

cos X sm X J */ 



852 INTEGRAL CALCULUS 



/• , • _- J Bin « COB* « . Bin X cob's . 1 . , . v . #• 

1. I sin^ X COB* xcix = h + — (Bin x cob x -f «) + C7. 

J 6 24 10 

■ 

SolMiion, FirBt applying formula [F], we get 

(A) JBin^x cos^xdx = h - Jcoa* xdx. 

[Herem=2, na4.] 

Applying formula [^] to the integral in the second member of (J.), we get 

yin /• . , sin X cos* X . 3 f , , 

(B) I coB^xdx = H - I coB^xdx. 

CHere m = 0, n B 4.] 

Applying formula \K\ to the second member of (B) gives 

sin X COB X X 



,r^ /• „ , smxcoBX . X 
(C) Jco8»xdx = + - 



Now BubBtitute the result (<7) in (^, and then thiB result In (^i). This gives the 
answer as above. 

- C . A o J C08X/8in*x sin'x sinxx . x . ^ 

2. I sm*x cos^ xdx = ( ) H h C. 

J 2\3 12 8/16 

3. I = tan X — 2 cot X — cot» x + C. 

J sin^xcos^x 3 

/•cos* xdx cotx.« « 3x ^ 

4. I = - -11^(3 - coB»x) - — + C. 
J BinSx 2 ^ '2 

. /•C08*oda COB a 3, ^ o - 

6. I — -— — = -r-^ COB a log tan - + C 

J Bin'a 28inao 1 ^ 2 

/. - , COB o /sin* a ,5., .5, \.6a.^ 
sin* oda = ( sin* a + - sin o I H yC, 
2\3 12 8 /16 



6 



^ r ^ cobB/ 1 . 3 \ 3, , tf ^ 

9. I C0b8 tdt = I COB^ « + - COB* < + — cos* < + — COB O + h C^« 

J 8\ 6 24 16 /128 

10. f ^?? = L_(^_L. + _J 5siny^ 

•/ sin* ^ cos* y co8*y\3Bin*y 3sin^ 2 / 

5 
+ - log(Bec y + tan y) + C 



REDUCTION FORMULAS 858 



cosnx 
n 



207. To find C^^ sin nxdx and C^^ cob nxda>. 

Integrating e"* sin nxdx by parts, 
letting w = e"', and dv = sin nxdx; 

then du = ae*^dx^ and t;= — 

« 

Substituting in formula (-4), p. 841, 
namely, 

we get 

/ >i\ r ax ' J ^"^ cos wx . a /* „ , 

(-4) I e" sm wzaa: = h - I « cos nxos; 

•/ n nJ 

Integrating «** sin n2:da; again by parts, 
letting u = sin nxy and dv = e'^dx; 

then c7if = n cos nxdx^ and v = — • 

a 

Substituting in (A)^ p. 341, we get 

{B) J «"* sin warda: = ^ I e"* cos nxdx. 

Eliminating I e"* cos nxdx between {A) and (5), we have 

(a^ 4-71*) I g" sin nxdx =. e'^{a sin nx — n cos na;), 

/' . , e'^ia sin na: — w cos na;) ^ 
e*" sm narda; = — ^^ '- -f C. 
or -f n* 

Similarly we may obtain 

/' , 6"'(n sin na; 4- a cos na;) . ^ 
«"* cos wa;ia; = — ^^ , ^ ^ '- 4- C. 

In working out the examples which follow, the student is advised 
not to use the above results as formulas, but to follow the method 
by which they were obtained. 



854 INTEGRAL CALCULUS 



e* COB xdx = — (sin x + cos x) + C7. 

2 
3. fe^'cosSxdx = — (Sein 3x + 2 cob3x) + O. 

. /"sin xdx sin X + cos x . ^ 

4. I = ^. c. 

6. I = — — -(28in2x-3coB2x) + C. 

J c»' 13c»* 

A r • • « ^ ^/t 2 8in2x + co8 2x\ . -, 

6. I c*8in* xdx = — ( 1 ^ ) + C. 

n r « J e«/- . 2sm2o + cos2o\ . -- 

7. J c*cos9ada = — f 1 + -^^ j + 0. 

a rc^co8-dx = e«^8in- + coe5^ + C. 
J 2 • \ 2 2/ 



c^^sinott 



9. I eo^(8Uiaa + cosaa)(ia = h C 

10. reS''(Bin2x-C08 2x)(ix= -^(8in2x ~6cob2s) + C. 
•/ 13 



CHAPTER XXIX 




--a-4 M N 



THE DBFINITB DTTBGRAL 

208. Differential of an area. Consider the continuous functioA 
^(a?), and let 

be the equation of the curve AB, Let 
CD be a fixed and MP a variable ordi- 
nate, and let u be the measure of the 
area CMP J).* When x takes on a suf- 
ficiently small increment Ax, u takes 
on an increment At^ (=:area MNQP). 
Completing the rectangles MNRP and MNQS^ we see that 

area MNRP < area MNQP < area MNQS, 

or, -MP.Ax< Aw<i\rQ.Ax; 

and, dividing by Ao;, 

Al£ 

Now let Ax approach zero as a limit; then since MP remains 
fixed and NQ approaches MP as a limit (since y is a continuous 
function of x), we get 



du 



= V(=MP), 



or, using diflferentials, 



dx 

du = ffdx. 



Theorem. ITie differential of the area bounded by any curve^ 
the axis of X^ and two ordinates is equal to the product of the ordi- 
nate terminating the area and the differential of the corresponding 
abscissa. 

• We may suppose this area to be generated by a yariable ordinate starting out from CD and 
moTing to the right ; hence « will be a function of x which yanishes when x^a. 

t In this figure MP is less than NQ; if MP happens to be greater than NQ^ simply reverse 
the inequality signs. 

865 



866 INTEGRAL CALCULUS 

309. The definite integral. It follows iiom the tlieorem in the 
last section that if AB is the locus of 

then du = ydr, or, 

(A) du=^{x)dx, 

where du is the differential of the area 
between the curve, the axis of x, and 
any two ordinates. Integrating {A)^ 
we get 

w = \^{x)dx. 

Since \^{x)dx exists (it ia here represented geometrically as an 
area), denote it hyf(x)+ C. 

(B) .■.u=f(T) + C. 

We may determine C, as in Chapter XXV, if we know the value 
of u for some value of j^ If we agree to reckon the area from the 
axis of y, i.e. when 

(C) x = a, u = eLre&OCDG, 
and when x=b, u = area OEFG, etc., 
it follows that if 

(D) x=% then m = 0. 
Substituting {D) in {B), we get 

w=/(0)+C, or C = -/(0). 

Hence from (B) we obtain 

{E) u=f(z)-f{0), 

giving the area from the axis of y to any ordinate (as MP). 

To find the area between the ordinates CD and EF, substitute 
the values (C) in {E), giving 

{F) area OCDG =f{a) -/(O), 

(Q) area OEFG =/(6) -/(O). 

Subtracting (F) from (G), 

(H) area CEFD = f{b) - /(«).■ 

> The itHdent 
fnticllOD ifUcIi d: 



THE DEFINITE INTEGRAL 357 

Theorem. The difference of the value% of j ydx for a; = a and 

x=b gives the area bounded hy the curve whose ordinate is tfy the 
axis of Xy and the ordinates corresponding to x=a and x = b. 
This difference is represented by the symbol* 



(/)" I ydx, or i (f>(x)dx, 

•/o •/a 



and is read " the integral from a to ( of ydxJ*^ The operation is 
called integration between limits, a being tlie lower and b the upper 
limit.f 

Since (J) always has a definite value, it is called a definite integral. 
For, if 

f<f>{x)dx=f{x)-^C, 

then £<t>{x)dx = [f(x)-^c][=[f(b) + c]-^[f(a) + c], . 

or, £<l>{x)dx=f{b)^f{a), 

the constant of integration having disappeared. 
We may accordingly define the symbol 

I <f>(x)dx 

as the numerical measure of the area bounded by the curve y =^<\> (:r), J 
the axis of X, and the ordinates of the curve at x=:a, x=b. This 
definition presupposes that these lines bound an area, i,e. the curve 
does not rise or fall to infinity, and both a and b are finite. 

We have shown that the numerical value of the definite integral 
is always /(6) —f{a), but we shall see in Ex. 2, p. 363, that/(6) —f{a) 
may be a number when the definite integral has no meaning. 

210. Geometrical representation of an integral. In the last section 
we represented the definite integral as an area. This does not 
necessarily mean that every integral is an area, for the physical 
interpretation of the result depends on the nature of the quantities 

* This notation Is due to Joseph Fourier (1768-1830). 

t The word limit In this connection means merely the value of the varlahle at one end of 
its range (end value), and should not he confused with the meaning of the word in the Theory 
of Limits. 

$ ^ (a;) is oontinaouB and single-ralued throughout the interral [a, b]. 



858 



INTEGRAL CALCULUS 



represented by the abscissa and the ordinate. Thus, if x and y 
are considered as simply the coordinates of a point and nothing 
more, then the integral is indeed an area. But suppose the ordi- 
nate represents the speed of a moving point, and the corresponding 
abscissa the time at which the point has that speed; then the 
graph is the speed curve of the motion, and the area under it 
and between any two ordinates will represent the distance passed 
through in the corresponding interval of time (see § 187). That 
is, the number which denotes the area equals the number which 
denotes the distance (or value of the integral). 

Similarly a definite integral standing for volume, surface, mass, 
force, etc., may be represented geometrically by an area. On page 
872 the algebraic sign of an area is interpreted. 

211. Mean value of f(x). This is defined as follows : 



r 








Q^ 




L 


n 


^^^^^^ 


^r 




} 


/ 


^ 




M 




A 




c 




B 





ar* 


a 




Z' 


"h 



sy 



Mean value of f(x) ^ j^r^^)^ 
from a5 = atoa5 = 6)"" 6 — a 

Since from the figure 

J <l>(x)dxz= area AFQB, 

this means that if we construct on the base AB(=b^a) a rec- 
tangle (as ALMB) whose area equals the area of AFQB, then 

, 2iVQ2i,ALMB AB'CR ,^.^ , ^^ 
mean value = — = — — = altitude CB. 



b — a 



AB 



212. Interchange of limits. 

Since J <\>{x)dx=f{b)—f(a)^ and 



we have 



£<t>{x)dx=f(a)-f(b) [/(6)-/(a)], . 

j ^{x)dx=—i ^{x)dx. 



Theorem. Interchanging the limits is equivalent to changing the 
sign of the definite integral. 



THE DEFINITE INTEGRAL 359 

213. Decomposition of the interral of integratloa of the definite 
int^ral. 

Since J<(>{x)dx=/(x{)~f{a) and 

f,i>{x)dx^f{b)-f{x;), 

we get by addition, 

fy{x)dx+f4.{x)dx=f(b)-f(a). 

But j^<f, (x) dx =f{b) -f{a) ; 

therefore by comparing the last two expressions we obtain 

Interpreting this theorem geometrically, as in § 209, p. 356, we 
see that the integral on the left-hand side represents the whole 
area CEFD, the first integral on the 
right-hand side the area CMPD, and 
the second integral on the right-hand 
side the area MEFP. The truth of the 
theorem is therefore obvious. 

Even if x^ does not lie in the interval 
between a and h, the truth of the theo- 
rem is apparent when the sign (p. 372) as well as the magnitude 
of the areas is taken into account. Evidently the definite integral 
may be decomposed into any number of separate definite integrals 
in this way. 

214. The definite integral a function of its limits. 

From £4,(x)dx^f(b)-f{a) 

we see that the definite integral is a function of its limits. Thus 
I ^ (z) dz has precisely the same value as I <ft(x)dx. 

Theorem. A definite irUegral is a Junction of its limitt. 



860 INTEGRAL CALCULUS 

215. Calculation of a definite integral. The process may be sum- 
marized as follows : 

First step. , Find the indefinite integral of the given differential 
expression. 

Second step. Svbstitute in this indefinite integral first the upper 
limit and then the lower limit for the variable^ and subtract the last 
result from the first. 

It is not necessary to bring in the constant of integration, since 
it always disappears in subtracting. 

Ex. 1. Find r z^dx. 

1 L3 J 1 3 3 

Ex. 2. Find f Bin xdz. 
Jo 

Solution. C Bin xdx = f - cob x 1 = I — (- 1) I — — 1 = 2. Aia, 



,'* dx 
Ex. 3. Find ' 



J'" ox 
a« + 



x« 



J"" dx rl x"i *• 1 1 
= - arc tan - = - arc tan 1 — arc tan 
a*-* + x^ La a J o a a 



*" ^ IT . 

= = — jing, 

4a 4a 



216. Infinite limits. So far the limits of the integral have been 
assumed as finite. Even in elementary work, however, it is some- 
times desirable to remove this restriction and to consider integrals 
with infinite limits. This is possible in certain cases by making 
use of the following definitions. 

When the upper limit is infinite, 

and when the lower limit is infinite, 

£j> (X) dx = „ ^^X fj (^) <i^ 
provided the limits exist. 



THE DEFINITE INTEGRAL 



861 



Ex. 1. Find f^*^. 

Solution. r^= li^^J' r^^^ limit r^ll* 

Ex. 2. Find I . 



5oZu<io7i 



^"^ i 



+• Sa*dz 



+* 8a«dg ^ limit f^ Sa^cte __ Umit 



X 1* 
" arc tan — 

2aJo 



= .^^™!^^r4aaarctanA"|=:4aa.!: = 2Ta». ulna. 
= +QoL 2oJ 2 



Let OS interpret this result geometrically. The graph of our function is the 

witch, the locus of 

8a8 
y =z . 

x«-+-4a« 

Area 0Pe6= f^-^^^^ = 4 a^ are tan A . 
Jox^ + ^a^ 2a 

Now as the ordinate Q6 moves indefi- 
nitely to the right, 

& 
4 a^ arc tan 

is always finite, and 

limit r4a2aretanA] = 2Ta2, 
6 = +ooL 2aJ 



2a 
6 




>4-ao 



which is also finite. In such cases we 

call the resuU the area bounded by the curve^ the ordinate OP, and OX, although 

strictly speaking this area is not completely hounded. 



Ex. 3. Find C 



+* dx 

X 



SoltUion. 



+ «dx 



/•+«ax__ limit r'*ax_ limit ..^„j^. 
Jl 7""& = + ooJi •^-6 = +oo^^^«^^- 



The limit of log & as 6 increases without limit does not exist ; hence the integral 
has in this case no meaning. 

217. When ^(o?) is discontinuous. Let us now consider cases 
when the function to be integrated is discontinuous for isolated 
values of the variable lying within the limits of integration. 

Consider first the case where the function to be integrated is 
continuous for all values of x between the limits a and ( except 
x = a. 



862 INTEGRAL CALCULUS 

If a < 6 and e is positive, we use the definition 
(A) f4>ix)dx=^^^lf<l>{x)dx, 

and when </> (x) is continuous except at a: = 6, we use the definition 

(B) fj (x) dx = ;'^'^ jr'" V ix) dx, 

provided the limits are definite quantities. 

Ex. 1. Find C 



dz 



Vaa-x« 



Solution. Here — — =: becomes infinite for x = a. Therefore, by (B), 

r «_^_ ^ limit r— _dx_ ^ limit T^^ ^j^ xl — 
^0 Va2 - x2 « = 0*^o Va^ - x^ « = 0L aJo 

= ]^™>^rarc8ln(l-')] = arc8inl = ^. ^n«. 

J •Idas 
x2 

Solution. Here — becomes Infinite for x = 0. Therefore, by {A), 

x* 

Jo x^ e = Ojc x2 e = 0\« / 
In this case there is no limit and therefore the integral does not exist 

If c lies between a and b and <f>{x) is continuous except at 2;=:<?, 
then, € and e' being positive numbers, the integral between a and J 
is defined by 

(C) £<l>{x)dx='^^'l£"4>(z)dx + ^''^l£j(x)dx, 

provided each separate limit is a definite quantity. 



Ex. 1. Find 



/•»« 2xdx 



(x2 - a2)i 

Solution, Here the function to be integrated becomes infinite for x = a, i.e. for 
a value of x between the limits of integration and 8 a. Hence the above defini- 
tion (C) must be employed. Thus, 

/•««_2xdx_ _ limit r°-* 2xdx limit /•**• 2xdx 

=«^'^[3(.-a,»];-v»-"o[«(*'-«')»]::.. 

= J*So[3-^(a-.)=-a« + 8a']+ ^'^'* [S-I^^ - 3 v'(a + 0«-«'] 
= 3o' + 6a' = 9o*. An$. 



TUE DEFINITE INTEGRAL 



To interpret thU geometrically, let us plot the gnph, i.e. tbe locua of 




= 3V(o-«}*-a* + 3a'. 
Now u PE movee to the right tonards 
the asymptote, Le. m « approache* zero, 




is always finite, and 



™i[a^ 



which U also flaite. As in Ex. 2, [ 
the asymptote, aod OX. Similarly, 



S61, 8 a* is called the area bounded by OP, 



nE'QRO 






is always finite as QE" moves to the left towards the asymptote, and as ^ approaches 
zero the result 6 a' is also finite. Hence 6 a' is called the area between QR, the 
asymptote, the ordinate x = Sa, and OX. Adding these results we get 9a}, which 
is then called tbe area to the right of OT between the curve, t^e ordinate x s 3a, 
and OX. 



Ex. 2. Find 



/■''' dx 
Jo (»-a)a' 



Solution. This fancllon also becomes infinite between the Umita of integraUon. 
Hence, by (C), ^ ^ ,,_^,^ ^ ^^^^ d, 

Jo (x-a)» » = 0j) (i-o)' •' = OJ-+.'(J!-a)» 
^ _ llmitr 1 T- ■ limit T 1 1*" 

_liniit/l 1\ , limit/ 1 . 1\ 

In this case tbe limits do not exist and the 
integral has no meaning. 

If we plot the graph ol this function and note 

the limits, the condition of things appean very 

much the same as in the last example. It turns 

out, however, that the shaded portion cannot be 

properly spoken of as an are&, and the integral sign has no meaning in this case. 

That It Is important to note whether or not the given function becomes Infinite 
within the limits of Integration will appear at once If we apply our Inlegtation 
formula without any investigation. Thus, 



Jt ix-a)* L !r-aj« 



a result which ia absurd In view of the above discnssious. 



864 



INTEGRAL CALCULUS 



1. r*6x%ix = 38. 

2. r (a«x-x«)dx = - 



3. r^=i. 

'(2x 



J "ax 
— = L 
1 X 

5. C^ (xa - 2x + 2)(x - l)(ix = 
•^° V3-2X 



-f 



.2x«dx 8 



7. j - = --log3. 

Jo X + 1 3 



dz 



V2 - 3x2 4V3 



9. f' f^ =-^126. 
2-v^?^ 

Jo yS - y + 1 3 V3 

r^ tdt _ log 2 
' J2 1 + ia - 2 

Jl X2(l-f 



+ * dx 

+ 



= l-log2. 



Jo a^ + x^ 2a' 

ji «(i+«)^ '^ 2 



5. r* sin 0d^ = 1. 

J*2ir 



er8aftW = 3x. 



7. p8ec*^(W = f. 

J '•I T 
arc sin xdx = 1. 
2 

L Jj'xlog 



xdx = 



e^ + l 



20. r e-«dx = l. 

21. f arc tan xdx = - — log Vi. 
Jo 4 

/•s*" V2r 

22. J -4r-dx = 4r. 



vs 



23 



• j^*(fVt-A«»)d« = 2V6-6. 



24. r /^ =^. 

26. r^;^=8r. 

•^0 V2r-y 



256 IT 6* 



27. 2a r (2 + 2co8^)*d5 = 8a, 



28 



. r* sin* a cos* ada = ^^j. 



29. r*^tanoda = 0. 

30. r log ydy = - 1. 
3L J xlogxdx = — J. 

82. j;'xnog«d« = -t. 



33 



•X 



a^ sin ada = ir — 2. 



34. r^8ec^ = log(?-tp:-?) 

6 

^0 2 + cos a 3 V3 



36 



J '2 cos Q^ __ X 
l + 8in2tf ~4 



THE DEFINITE INTEGRAL 865 

218. Change in limits corresponding to change in the variable. 
When integrating in Chapter XXVII by the substitution of a new 
variable it was often found quite troublesome to translate our result 
back into the original variable. When integrating between limits, 
however, we may avoid the restoration of the original variable by 
changing the limits to correspond with the new variable.* Ttiis 
process will now be illustrated by an example. 

Ex. 1. Find C Va* — x' (2x, assuming x = asin $. 



Solution. Va^-x^dx = a* Vl - Bin^ e . cos Bde = o«co8« 6d$. 
When x = a, a = a^in9f i.e. 9 = -; 

and when x = 0, = a sin 9, i.e. $ = 0. 

.•.j(;'VS?T^<te=j;i«*co8«ftl»=[|(» + !!!^)]» = ^. An,. 

By this change in limi^ we mean that as $ increases from to - « x will increase 
from to a. 

EXAMPLES 

1. r - = 4-21og3. Assume Vx = «. 

•^M + Vx 



a dx 



Assume x^az. 



2. f ^ „ 

8. r'<^-)^ = 6. . A««me* = l. 

Ji X* z 

4. r*__^_ = 1. Assume VT^ = 2. 
•^0 Vl -x« 

5. r*sinaco8«o(ia = i. Assume sin a = «. 

g r?(Bin ^H- cos g)(i^^ logs Assumesin^-cos^c=z. 

Jo 3 + sin 2^ 4 

7. I = arc tan c — - • Assume c* = «. 

Joe*-\-e-' 4 

8. r°_ ^g — _ ^ Assume x = o sln*«. 
•'o Vox-xa 

• The relation between the old and the new variable should be euch that to each ralue of 
one within the limits of integration there is always one, and only one, finite ralue of the other. 
When one is given as a many-vained function of the other, care must be taken to choose the 
right values. 



366 INTEGRAL CALCULUS 

3V3 



9. r<^-"i'^=8+ 

J^ (ai-2)J + 3 



Aasame x — 2 = z*. 



— dx = 4 — r. Assume e* - 1 = z*. 

c* + 3 

1. r -J^_ = i^. Assume 2 + 4y = z«. 

^i V2 + 4y 2 

Assume 2 = tan - • 



2 cos t V5 2 



Jo 3 + 

L r — dx = 4 — T. Assume c* — 1 = «*. 

Jo c* + 3 

, /•« x^dx 8V2-4 . 

L I 5 = a. Assume x = a tan 2. 

Jo (a2 + x*)* 2 

J* * y irci* 

y2 Va* — y^dy = — . Assume y = a sin z. 

*" '^ *" 16 

6. r V2« + t2(tt = V3- Jlog(2+V3). Assumet + 1 = «. 

J«»l0J« 4 — IT 
V c* — 1 dx = Assume e* + 1 = «. 

2 

8. I — do = - log 3. Assume sm — cos = z, 

Jo 3 + sin2^ 4 

/.i+v^ (x» + l)dx , . .1 

9. I — = log 3. Assume x = «. 

•^^ xVx*4-7xa + l * 



CHAPTER XXX 



INTEGRATION A PROCESS OF SUMMATION 



219. Introduction. Thus far we have defined integration as the 
inverse of differentiation. In a great many of the applications of 
the Integral Calculus, however, it is preferable to define Integra^ 
tion as a process of summation. In fact the Integral Calculus was 
invented in the attempt to calculate the area bounded by curves 
by supposing the given area to be divided up into an "infinite 
number of infinitesimal parts called elementSy the sum of all these 
elements being the area required." Historically the integral sign 
is merely the long /S, used by early writers to indicate " sum," 

This new definition, as amplified in the next section, is of fun- 
damental importance, and it is essential that the student should 
thoroughly understand what is meant in order to be able to apply 
the Integral Calculus to practical problems. 

220. The definite integral defined as the limit of a sum of differ- 
ential expressions. Assume (f>{x)dx as the differential of f{x). 
Then by § 209, p. 366, 



(^) 



f<l>{x)dx=f{b)^f{a) 



gives the area bounded by the curve y = <f>{x) (AB in figure), the 

axis of X, and the ordinates 2; = a and x = b.* Now suppose the 

segment ai to be divided into a number of equal parts, say 6, each 

equal to Ax^ at points whose 

abscissas are b^^ b^ 6„ b^^ b^. 

Erect the ordinates at these 

points and apply the Theorem 

of Mean Value (44), p. 168, to 

each interval, noting that here 

<\>{x) takes the place of 4>\o^, 




r 



• In the figure a, 6^ &,, x,, Xfi, etc., denote the abfclMM of the points nnder whidh thej Are 
written. 

867 



368 INTEGRAL CALCULUS 

Applying (44) to the first interval (a = a, 6 = 6i, and x^ lies between 
a and b^ as shown in figure), we have 

0^ — a 

or, since 6^ — a = Aa:, 

f{b;)-f{a)=<f>{x,)Ax. 

Applying (44) in the same way to each one of the remaining 
five segments, we get 

/(!>*) -f{f>^=4>{^.)^ 
f{K)-nh)=H^.)i^ 

m)-m)=<^(^.)^ 

. , f{b)-f{b,)=4>(x,)Ax, 

respectively. 

Adding these six equations, 

(B) f{b) -/(a) = </> (X,) Aa: + </> (x,) Ax + </> (x,) Ax 

+ ^ (ar J Aa: 4- </) (Xg) A2: + 4>{x^)Ax. 

But <l>{x^)Ax = area of first rectangle, 

^(z2)Ax= area of second rectangle, etc. 

Hence the sum on the right-hand side of (B) equals the area of 
the figure bounded by the zigzag line BF^p^P^p^^p^P^p^F^p^Q^ 
the axis of Xj and the ordinates x=:a and 2; = (. It is also evident 
that the area of this figure equals 

f(f>) -/(«), 
no matter into how many equal parts the interval [a, 6] may be 
divided. Hence for any number n of equal parts 

(0) f{b)^f{a) = <l>{x,)Ax-^<f>{x,)Ax-^...+<f>{x,)Ax, 

IB) where Ax=z "" . 

n 

When the number of segments (=n) into which the interval 
[a, 6] is divided increases without limit, equations {(T) and (2>) still 
hold true, but Ax becomes dx^ i.e. a variable whose limit is zero 
(§30, p. 21). 



INTEGRATION A PROCESS OF SUMMATION 869 

.-. /(J) ~f{a) = ^ta'^ [.^ (^,) i:r + ^ (^ ir + . - . + ^ (^.) dx}, 
or, by {A), 

(E) £4>{x)dx^^^fl{.^{x,)dT+4,{x,)dx^.-. + <^{x,)dx\ 

This exhibits our definite integral as the limit of a sum of differ- 
ential expresaiona. Each one of the differential expressionB ^(a',)(2x, 
■ ■■t 4* (^i) ^^ '3 called an element of the integral. 

The above result also illustrates very clearly the definition of 
the definite integral as the area under the curve. For as n increases 
without limit the sum 

4,(x^)dx + <t){x;)dx + ---+<f>(xjdx 
always represents the area under the zigzag line, and evidently 
the figure bounded by the zigzag line, the end ordinates, and OX 
approaches the area under the curve as a limit. 

We may, however, attack the problem in a much more general 
way, for it is geometrically evident that a Bubdivieion of the given 
area may be made in an infinite variety of ways such that the con- 
tinuation of the process will lead in the limit to the area desired. 
For example, let us choose within the interval [a, 6], n — 1 abscissas, 
Zi, x^ '■-, x^_^, in any manner whatever, and erect ordinates at the 
corresponding points to the curve. Then the area is divided into 
n portions such as x^P, x^Q, ete. Denote the lengths of the sub- 
divisions on OX as follows : 
x,-a = Ax^, 
Xt-Xi = Aa^ 

a:,_, — a:,_, = Ax,_„ 
h — a:,_, = Ai.. 
Then evidently the area of 
the segment x^Q, for example, equals approximately that of the 
rectangle whose base is Az^ and altitude Qx„ or (x,). Carrying 
out this idea for each portion, we have as a result that 

*(ar,)Ar, + 0(s,)Az, + ..-4.0{i._,)Ax„, + 0(OA3:J 
gives an area approximately equal to the area required. And now 
it is geometrically obvious that the limit of the above sum is the 



370 INTEGRAL CALCULUS 

area under the curve if the process of subdivision of [a, b] be 
continued according to some law by which each division on x 
approaches the limit zero. That is, 



^=lX<t>i^d^,*=£4>(=ddx. 



This general discussion shows that inteffration between limits is 
a process of summation^ in which, however, the definite integral 
appears not as a simple sum but as the limit of a sum^ the number 
of terms increasing without limits and each term separately approach- 
ing the limit zero. 

In order to replace the intuitional point of view that we have so fax adopted in 
the text by a rigorous and general analytical proof, proceed as follows. 

Divide the interval [a, 6] into any number of parts Axi, Axs, • -*, Az., and let 
vfi be any abscissa in the segment Axi, the extremities of this segment being denoted 
by X{ and Xi^ i. Also suppose Xj to be a value in the interval [x„ Xi^i\ determined 
by the Theorem of Mean Value (as in the first part of this section), so that 

/(X.+ 1) - /(Xj) = (x;) Ax,-. 

Then, exactly as hrfoTBy the sum 

equals the required area. And while the corresponding sum 

n 

(O) ^^{x'i)Axi 

1 
does not also give the area, nevertheless we may show that the two sums (F) and (G) 
approach equality when n increases without limit. For the difference 4>{xi) — (z'l) 
does not exceed in numerical value the difference of the greatest and smallest 
ordinates in Ax^. And furthermore, it is always possible t to make all these differ- 
ences less in numerical value than any assignable positive number e, however small, 
by continuing the process of subdivision far enough, Le. by choosing n sufficiently 
large. Hence for such a choice of n the difference of the sums (F) and {G) is less 
in numerical value than e (& — a), i.e. less than any assignable positive quantity, 
however small. Accordingly as n increases without limit, the sums (F) and (O) 
approach equality, and since (F) is always equal to the area, the fundamental result 
follows that ^ n 

in which the interval [a, b] is subdivided in any manner whatever, and x'j is any 
abscissa in the corresponding subdivision. 

* Any term of this sum is Bometlmee called an element of the area, for each one represents 
the area of one of the rectangles forming the whole figure. 

t That such is the case is shown In advanced works on the Calculus. 



INTEGRATION A PROCESS OF SUMMATION 



871 



The process of evaluation of the limit of this sum is accordingly 
often spoken of as summing up an infinite number of infinitely sm^ill 
quantities^ but the phrase has no meaning except in the above sense. 

We may apply to great advantage this fruitful idea of summing 
up an infinite number of infinitely small quantities to a large number 
of the problems of the Integral Calculus. In order to obtain solu- 
tions by this method the following steps are in general to be taken. 

First step. Find a differential expression for any one of the infin- 
itesimal quantities (elements) composing the quantity to be calcu- 
lated,* and reduce it to a form involving only a single variable. 

Second step. Integrate this differential expression (i.e. sum up 
all the elements) between the limits given by the conditions in the 
problem. 

221. Areas of plane curves. Rectangular coordinates. It was 
shown in § 209, p. 856, that the area between a curve, the axis 
of X and the ordinates x = a and x=b 
is given by the formula 



U) 



area 



= J ydXi 







the value of y in terms of x being sub- 
stituted from the equation of the curve. 
Considered as a process of summation, 
it is customary to look upon this operation as follows (see p. 867). 
Consider any strip (as CE) as an element of the area. Regard- 
ing it as a rectangle of altitude y and infinitesimal base dx, its area 
is equal to ycZx, and summing up all such strips between the ordi- 
nates AF and BQ gives the area ABQP. 

Similarly it may be shown that the 
area between a curve, the axis of F, and 
the lines y = c and y = rf is given by the 
formula 




(B) 



area 



=X' 



xdy^ 



the value of x in terms of y being substi- 
tuted from the equation of the curve. 



* If an area la wanted, find /i differential exprecMion for an element of the area ; if a length, 
fled it for an element of the length ; if a volume, find it for an element of the Tolume, etc. 



372 



INTEGRAL CALCULUS 



Ex. 1. Find the area included between the semicubical parabola y< = x* and 
the line x = 4. 

SoliUioTL Let us first find the area OMP^ half of the required area OPP'. For 
the upper branch of the curve y = Vx", and summing up all the strips between 

the limits x = and x = 4, we get, by substituting in (A), 



/ 




area OMP =Cydx -Cz^dx = V = ^^' 



Hence area OPP" = 2 . V = 26§. 

If the unit of length is one inch, the area of OPP is 
^^X 26 J square inches. 

Note. For the lower branch y = — x* ; hence 
area OMP" = r(- x^) dx = - 12f . 

This area lies below the axis of z and has a negative sign 
because the ordinates are negative. 
In finding the area OMP above, the result tooa positive because the ordinates 
were positive, the area lying above the axis of x. 

The above result, 25 j^, was the total area regardless of sign. As we shall illus- 
trate in the next example, it is important to note the sign of the area when the 
curve crosses the axis of X within the limits of integration. 



Ex. 2. Find the area of one arch of the sine Y 
curve y = sin x. 

Solution. Placing ^ = and solving for x, 

X = 0, IT, 2 IT, etc. 
Substituting in (A), p. 371, 



we find 




Also, 



and 



area OAB = C ydz = j sin xdx = 2. 
area BCD = C ydz = ( 'sin xdx = — 2, 
area OABCD -C ydz = f 'shi xdx = 0. 



This last result takes into account the signs of the two separate areas composing 
the whole. The total area regardless of these signs equals 4. 



Ex. 3. Find the area included between the parabola x^ = 4 ay and the witch 

8o» 



y = 



x2 + 4 a2 



Solution. To determine the limits of integration, we solve the equations simul- 
taneously to find where the curves intersect. The coordinates of A are found to 
be (-2 a, a), and of C (2 a, a). 



INTEGRATION A PROCESS OF SUMMATION 



373 



It is Been from the figure that 





area ^OCB 


Bat 


BxeADECBA 


and 


KteskDECOA 


Hence 


«re&AOCB 



ATeskDECBA - areaDJSCO^. 

•»« SaMx 



= 2Ta' 

xa + 4a« 



2«x2 , 4a2 



= 2 X area 0-EC = 2 f — dx = 

Jo 4a 

= 2ira« - — = 2a2(T - f). Ans, 



Another method is to consider the strip PS as an element of the area. If y' is 
the ordinate corresponding to the witch, and y^' to the parabola, the differential 
expression for the area of the strip PS equals (y^ — y'^dz. Substituting the values 
of 2^ and y'' in terms of z from the given 
equations, we get Y 



axeskAOCB = 2 x area OCB 

= 2rV-y")<^ 

Jo \x2 + 4a« 4a/ 
= 2a»(T~i). 



Ex. 4. Find the area of the ellipse — + ^ = 1. 





Solution. To find the area of the quad- 
rant OAB, the limits are x = 0, z = a ; and 

y = - Va2 - x2. 
a 

Hence, substituting in (^), p. 371, 

area 0-4B = - Cia^ - x2)*dx 
aJo 

= — (o^ — a;2)> + arc sin 
L2a^ '2 oJo 



ra6 



Therefore the entire area of the ellipse equals wab. 

222. Area when curve is given in parametric form, 
equation of the curve be given in the parametric form 

z=/(0, y = <f>{t). 



[B], p. 346. 



Let the 



874 INTEGRAL CALCULUS 

We then have i/=:<f>(t) and dx =f'{t) dt, 
which substituted* in (-4), p. 371, gives 

(A) area=: C*^(t)f^{t)dt^ 

where t = t^ when a? = a, and t = t^ when a; = J. 

We may employ this formula {A) when finding the area under a 
curve given in parametric form. Or we may find y and dx from 
the parametric equations of the curve in terms of t and dt and 
then substitute the results directly in (A)^ p. 371. 

Thus, in finding the area of the ellipse in Ex. 4, p. 373, it would have been 
simpler to use the parametric equations of the ellipse 

X = a cos 0, ^ = 6 sin 0, 

where the eccentric angle ^ is the parameter ($ 79, p. 94). 

Here y = 6 sin and dx = — a sin 0(2^. 

When X = 0, = - ; and when x = a, = 0. 

Substituting these in (A)^ p. 371, we get 

areaOJ.B=l ydx = — J a6sin*0<i0 = -— . 

1 

Hence the entire area equals rob, Ans, 

EXAMPLES 

1. Find the area bounded by the line y = 6x, the axis of X, and the ordinate 
x = 2. Am. 10. 

2. Find the area bounded by the parabola |/< = 4 x, the axis of X, and the lines 
X = 4 and x = 9. Ana, 2b\* 

3. Find the area bounded by the parabola y' = 4x, the axis of F, and the 
lines y = 4t and y = 6. Ana. 12}. 

4. Find the area of the circle x' + y* = r^. Ans. rr*. 

5. Find the area between the equilateral hyperbola xy = a>, the axis of X, 
and the ordinates x= a and x = 2 a. Ana. a31og2. 

6. Find the area between the curve y = 4 — x^ and the axis of X. Ana. 10]. 

7. Find the area intercepted between the coordinate axes and the parabola 
X* + y* = a*. 



s 



Ana. _. 
6 

• For a rigorouB proof of this satMtltutlon the student is referred to more advanced treatises 
on the Calculus. 



INTEGRATION A PROCESS OF SUMAIATION 375 

8. Find the area bounded by the semicubical parabola y' = z\ the axis of F, 
and the line y = 4. ^„^, J 4^1024. 

9. Find the area between the catenary y=- ^ + e ^ L the axis of F, and 
the line x = a. *^ "- - a*ro ^-. 

10. Find the area between the carve y = logx, the axis of JT, and the ordinates 
X = 1 and X = a. Ana, a (log a — 1) + 1. 

11. Find the entire area of the curve 



(ly- (D - '■ 



12. Find the entire area of the curve a*y* = x» (2 a — x). Ans. ira*. 

13. Find the area bounded by the curves 

X (y — e*) = sin X and 2 xy = 2 sin x + x*, 

the axis of F, and the ordinate x = 1. Ana. (^(e* — }x^dx = e— { = 1.56 + • • •. 

14. Find the area between the witch y = —„ — j— « smd the axis of X, its 

X* -f- 4 cfl 
asymptote. Ana. ^ra^, 

x* 

15. Find the area between the cissoid y* = ^ and its asymptote, the line 

X = 2 a. Ana. 3 ra^. 

16. Prove that the area bounded by a parabola and one of its double ordinates 
equals two thirds of the circumscribing rectangle having the double ordinate as 
one side. 

17. Find the area included between the two parabolas v^ = 2px and x> = 2py. 

Ana. — ^' 
8 

18. Find the area included between the parabola y^ = 2x and the circle 
y« = 4x-x2. Ana. 0.476. 

19. Find the total area included between the curve y = z^ and the line y = 2x. 

Ana. 2. 

20. Find an expression for the area bounded by the equilateral hyperbola 
z^ — y^ = a^, the axis of X, and the diameter through any point (x, y). 

An;a. — log • 

2 * a 

21. Find by integration the area of the triangle bounded by the axis of Y and 
the lines 2x + y + 8 = and y = — 4. Ana. 4. 

22. Find the area of the circle 

IX = r cos By 
y = rsintf; 
$ being the parameter. Ana. in4. 



876 



INTEGRAL CALCULUS 



23. Find the area of one arch of the cycloid 

(x = a(tf-8intf), 

\y = a(l — CO80); 
$ being the parameter. 

Hint. Since x vorieB from to 2 wa, varies from to 2 a-. 

Aris. 3 ra^ ; that is, three times the area of the generating circle. 

24. Find the area of the hypocycloid 

IX = o cos* Of 
y =:a sin* $ ; 



$ being the parameter. 

T 



Ara, 




8 
area of the circumscribing circle. 



; that is, three eighths of the 



25. Find the area of the loop of the folium of Descartes 
sc' + y» = 3 axy. 

3ai 



Bint. Let 



y^.tx; then X' 



l + /» 



1-f ^ 
The limits for t are and oo. 



y=l<!^,anddir-l:i2^3c««. 



{1 + !»)« 



223. Areas of plane curves. Polar coordinates. Let J^Cbeacurye 
whose equation is given in polar coordinates. Let (/>, 0) be the 
coordinates of P, and assume u as the measure of the area OEPJ* 
When takes on a small increment A5, 
u takes on the increment Am(= area OPQ). 
Completing the circular sectors OPR and 
OSQ^ it is seen that 



C F 



or 



area OPR < area OPQ < area OSQ, 

Dividing through by A0, 

Au 




Now let A0 approach zero as a limit ; then OQ will approach OP 
as a limit, and we get 

dd 2^\ 2 J 



* Since we may enppofle this area to be generated by a variable radliu reotor starting out 
from OE and moving up to the position OP, u will be a function of 9 which vanishes when 9s a. 

t The area of a circuliu' sector = i radius x arc= ^ OP x 0P^6. 

t In this figure OP is less than OQ ; if OP is greater than OQ, simply reverse the Inequality 
signs. 



INTEGRATION A PROCESS OF SUMMATION 377 

Or, using differentials, du= ^ p'dO, 
this being the differential of the area in polar codrdinatet. 
Integrating, we have u = ^ fp'dO. 

If we now apply the same line of reasoning as that followed in 
§ 209, p. 856, the area of the sector OEF may be calculated by 



s of the formula 



.=!jVd», 



the value of ^ in terms of being substi- 
tuted from the equation of the curve. 

To look at this process as a summation, 
consider any sector OBC as the element of 
area. Regarding it as a circular sector of 
radius p and infinitesimal arc pd0, its area 

equals ^ p*d0. Summing up all such sectors between OE and OF 
gives the area OBF. 

Formtda (A) may then he uted for finding the area hounded hy a 
polar curve and ike radii vectors correaponding to 6^a and = p. 

Ex. 1. Find the entire area of the lemniscate p' = <ii cos 2 8. 

Solution. Since the figure la symmetrical 
with respect to both OX and OF, the whole 
PiP,ir) area = 4 times the area of OAB. 

Since p = when ^ = j> ve see that it 9 

varies from to - • the radius vector OPsweeps 

OAB. Hence,snlMtitutingiD(^), 

'tkOAB = 4.iJ pW = 2a'J'*cqB2AM = o»; 

a of a square constructed on OA as 



1. Find the area swept over in one revolution by the radius vector of the spiral 
of Arcbiinedee, p = a$, starting with fl = 0. . 4 ir*a' 

3 

2. Find the area of one loop of the curve /> = ocoa 2 A ^^ ^. 




878 INTEGRAL CALCULUS 

3. Show that the entire area of the curve p = a sin 2 tf equals one half the area 
of the circumscribed circle. 

4. Find the entire area of the cardioid p = a (1 ~ cos ^). 

Ana. ; that is, six times the area of the generating circle. 

5. Find the area of the circle p = a cos tf. . xa* 

jn.n8, • 

4 

6. Prove that the area of the three loops of p = a sin 3 equals one fourth of 
the area of the eircumscribed circle. 

7. Prove that the area generated by the radius vector of the spiral p = e* equals 
one fourth of the area of the square described on the radius vector. 

8. Find the area of that part of the parabola p = a sec^ - which is intercepted 

between the curve and the latos rectum. . 8 a^ 

Ana, — . 
3 

9. Show that the area bounded by any two radii vectors of the hyperbolic 
spiral p0 = a ia proportional to the difference between the lengths of these radii. 

10. Find the area of the ellipse p8 = — — — — Ana, wab. 

'^ a2 sin* e-\-l^ cos2 

11. Find the entire area of the curve p = a (sin 2 + cos 2 0), Ana. ira\ 

12. Find the area of one loop of the curve p> cos 9 = a> sin 8 ^. «, « « 

- 3 a* o*. _ 
Ana, -— -T-log2. 



13. Find the area below OX within the curve p = a sin> - . 

Ana. (10ir + 27V3)^- 

224. Length of a curve. By the length of a straight line we 
commonly mean the number of times we can superpose upon 

« 

it another straight line employed as a unit of length, as when 
the carpenter measures the length of a board by making end-to- 
end applications of his foot rule. 
^^^^^^»w^^ Since it is impossible to make a 

j^^^ ^^\. straight line coincide with an arc of a 

Y ^1^ curve, we cannot measure curves in the 

I same manner as we measure straight 

lines. We proceed then as follows. 
Divide the curve (as AB) into any number of parts in any man- 
ner whatever (as at C, i>, E) and connect the adjacent points of 
division, forming chords (as AC, CD^ DE^ EB). 



INTEGRATION A PROCESS OF SUMMATION 



879 



The length of the curve is defined as the limit of the sum of the 
chords as the number of points of division increases without limit in 
such a way that at the same time each chord separately approaches 
zero as a limit. 

Since this limit will also be the measure of the length of some 
straight line, the finding of the length of a curve is also called 
"the rectification of the curve." 

The student has already made use of this definition for the 
length of a curve in his Geometry. Thus the circumference of a 
circle is defined as the limit of the perimeter of the inscribed (or 
circumscribed) regular polygon when the number of sides increases 
without limit. 

The method of the next section for finding the length of a plane 
curve is based on the above definition, and the student should note 
very carefully how it is applied. 

225. Lengths of plane curves. Rectangular coordinates. We shall 
now proceed to express in analytical form the definition of the last 
section. Given the curve 

y =m 

and the points P (a, c), Q (6,d) on it ; to find 
the length of the arc PQ. 

Take any number (= n) of points on the 
curve between P and Q, say P\ P'\ • •, P^*\ 
and draw the chords PP\ P'P'\ •••, P<">e. 
Consider any one of these chords, P'P"^ for example, and let the 
coordinates of P' and P" be 

P' (x\ y') and P" (x' + Ax\ y' + Ay'). 

Then, as in § 102, p. 141, 

P^P" = y/(Axy + {Ay')\ 




or. 



'--Mm'^- 



[DiTiding inBide the radical by (Ax')* and moltiplylng oatside by Aa^.] 

But from the Theorem of Mean Value, (42), p. 167 (if Ay' is 
denoted by f{b) -f(a) and Aa;' by J - a), we get 



^ =/■<->• 



x' <x^< x' ' 



380 INTEGRAL CALCULUS 

Subatituting, we get 

P'P" = [l+/'(2-i)«]*Ax'. 

In the same manner we find 

P'P" = [!+/' (xJ'J'Ax', 

• • • • • 

The length of the inscribed broken line joining P and Q (sum of 
the chords) is then the sum of these expressions, and the length 
of the arc P^ is therefore, by definition, the limit of the sum 

[1 +/' (xo)']^^'' + [1 +/' {x,r]^^' + ••• + [!+/' {^JT *Az^"^ 

as n increases without limit Hence, if we denote the length of the 
arc FQ by «, we have, by § 220, p. 367, the formula for the length of 

the arcy >.& 

«=J^[l4-/'(r)^*dx, or, 

where -~ must be found in terms of z from the equation of the 
ax 

given curve. 

Sometimes it is more convenient to use y as the independent 
variable. To derive a formula to cover this case, we know from 

(88), p. 162, that rf 1 rf 

-^ = -— ; hence ax = —- ay, 
ax ax ay 

Substituting this value of dx in (-4), and noting that the corre- 
sponding y limits are c and c2, we get* the formula for the length of 
the arC| , 



"» -rKg) +']*'• 



dx 
where — in terms of y must be found from the equation of the 

given curve. 



INTEGRATION A PROCESS OF SUMMATION 



381 



Ex. 1. Find the length of the circle z* + y^ = i^. 

Solution, Differentiating, — = — . 

dz y 

Substituting in (A)^ 

«.M=j;'[i+?)'d. 

r Substituting y«=» r«-a;« from the equation of the] 
[circle in order to get everything in terms of x.J 

.*. arc BA = r \ , = r arc sin - : 

Hence the total length equals 2 rr. Ans. 




XT 

2 



EXAMPLES 

1. Find the length of the arc of the semicubical parabola ajfl = x* from the 



origin to the ordinate x = 5 a. 



2. Find the entire length of the hypocycloid x' + y' = a'. 



Ana. 



335 a 



27 
Ana. Oo. 



X X 

3. Rectify the catenary y = - (c«+ c~") from x = to the point (x, y). 

^ a - -- 

Ana. -(e" — c «). 
2^ 



4. Find the length of one complete arch of the cycloid 

X = r arc vers - — V2 ry — y*. 

r 

dx y 



Ana. 8 r. 



Hint. U86(B). Here — = 



^y ^2ry-y* 

6. Find the length of the arc of the parabola y^ = 2jxc from the vertex to one 

extremity of the latus rectum. . p V2 . p , ,, . /iTv 

Ana. ^-r— +f log(l+v2). 

6. Rectify the curve oy* = x (x - 3 a)^ from x = to x = 3 a. Ana. 2 a Vs. 

7. Find the length in one quadrant of the curve (-) +(-) =1. 

^"^ ^''\ ffi + ab + V 



Ana, 



a-^b 



8. Find the length between x = a and x = 6 of the curve ev = 



Ana. log 






c8«-l 



+ a-6. 



9. The equations of the involute of a circle are 

tx = a (cos ^ + ^ sin $)y 
y = a (sin — cos 0), 

Find the length of the arc from ^ = to = ^i. 



Ana. \aO-^, 



382 INTEGRAL CALCULUS 

226. Lengths of plane curves. Polar coordinates. Formulas (A) 
and (B) of the last section for finding the lengths of curves whose 
equations are given in rectangular coordinates involved the differ- 
ential expressions 

[i.(g)-]'.»a[(|)Vx]W. 

In each case, if we introduce the differential of the independent 
variable inside the radical, they reduce to the form 

Let us now transform this expression into polar coordinates by 
means of the substitutions 

x = p cos 5, f/ = painO, 

Then dz=:z-^p sin Odd -f cos Odp^ 

and (iy = p cos 0d0 -f sin 0dp^ 

and we have 

[djc* + rfi/^*= [(- P sin ^d0 -f COS 0dpf 4- (p cos 0d0 + sin 0dp)^^ 
^[p*d0'-\^dp^]K 

If the equation of the curve is 

p ^m. 

then dp==f{0)d0 = ^d0. 

Substituting this in the above differential 
expression, we get 

If then o and fi are the limits of the independent variable 
corresponding to the limits in (A) and (B), p. 308, we get the 
formula for the length of the arc, 




(A) 

'a 



-S>^i%)i^ 



where p and ^ in terms of must be substituted from the equa- 

d0 
tion of the given curve. 



INTEGRATION A PROCESS OF SUMMATION 



883 



In case it is more convenient to use p as the independent vari- 
able, and the equation is in the foim 

then d6 = ^^(fi)dp = '—dp. 

dp 

Substituting this in [p^dff^ -f rfp']* 



Hence, if ^j and p, are the corresponding limits of the inde- 
pendent variable />, we get the formula for the length of the arc, 



(■B) 



dO . 



•=/,''[''(g)'+>]''"- 



where — in terms of p must be substituted from the equation of 
the given curve. 

Ex. 1. Find the perimeter of the cardioid p = a (1 + cos 0), 

dp 
SoltUion. Here — = — a sin ^. 

de 

If we let vary from to r, the point P wiU gener- 
ate one half of the curve. Substituting in {A), p. 882, 



- = C'[ai (1 + cos ey + a2 sina e(]^d0 
= a C'{2 + 2 cos tf)*(W = 2a f 'cos -de = 4a. 



a = 8 a. Ans, 




EXAMPLES 

1. Find the length of the spiral of Archimedes, p = aB^ from the origin to the end 

oi the first revolution. . ^/r~nr^ . ^i /o . ^/ , . ^ o v 

Ans. ira vl + 4t2 + -log(2ir + vl + 4Ta). 

2. Rectify the spiral p — tf^ from the origin to the point (p, ^). 



Hint. Ute(^. 



Aim. " Va» -f 1. 
a 







3. Find the entire length of the curve p = o sin*- 



Am, 



3ira 



4. Find the circumference of the circle p = 2 r sin ^. Ans. 2 irr. 

6. Find the length of the hyperbolic spiral pe=a from (pi, ^i) to (pj, ^2)- 

_ f 2 (a + Voa + pi«) 



k 



384 



INTEGRAL CALCULUS 



6. Find the length of the arc of the cisBoidp = 2a tan^ sln0 from = to = 



An8. 2a|V6-2-V81og ^"^"^ l 
t V2(2 + V3)J 



$ WW 

7. Find the length of the parabola p = a aec^- from tf = — -totf = -- 

2 2 ^ 



Ana. 2a(8ec— + logtan— Y 



a Show that the entire length of the epicycloid 4(p«- a^)' = 27a*p2sin«^, 



a 



which is traced by a point on a circle of radios - rolling on a fixed circle of radius 
a, is 12 d. 

227. Volumes of solids of revolution. Let V denote the volume 
of the solid generated by revolving the plane surface AMFC about 

the axis of X, the equation of 
the plane curve CFD being 

Let X (= 03f) take on a small 
increment Ax (= MN) ; then 
V takes on an increment A T, 
the volume of the disc gen- 
erated by the plane surface 
MNQP. In revolving, the 
two rectangles MNRP and 
MNQS generate cylinders having the same altitude Ax{-=^MN)y 
the exterior one having NQ^ and the interior one MP^ as radius 
of the base. The disc generated by MNPQ is evidently greater 
than the interior but less than the exterior cylinder. Hence 




irMF '£Lc<AV<TrNQ 'Ax\ 



or, dividing by Aa:, 



AF 



IT MP < -r- < irNQ . 

Ax 



Now let Ax approach zero as a limit ; then NQ approaches MF 
as a limit, and we get 

dV 



dx 



= vf(=7rMP); 



or, using differentials. 



dV= TT'ifdx^ 



INTEGRATION A PROCESS OF SUMMATION 



385 



which is the differential of the volume of the solid of revolution. 

Integrating, 

F= TT I f/^dx. 

Therefore if OA=a and ffB=zh^ the volume generated by revolv- 
ing ABDC about the axis of X may be calculated by means of the 
formula 



(J) 






where the value of y in terms of x must be substituted from the 
equation of the given curve. 

This formula is easily recalled if we consider a slice or disc of 
the solid between two planes perpendicular to the axis of revolu- 
tion as an element of the volume, and regard it as a cylinder of 
infinitesimal altitude dx and with a base of area iry\ hence of vol- 
ume iry^dx. Summing up all such slices (elements) from -4 to -B, we 
get the volume generated by revolving ABDC about the axis of X 

Similarly when F is the axis of revolution we use the formula 



(B) 



= IT j a>^dy, 



where the value of x in terms of y must be substituted from the 
equation of the given curve. 

Ex. 1. Find the volume generated by revolving the ellipse -^ + ^ = 1 about the 
axis of X. a 63 

Solution. Since y* = - (a^ - a;«), and 

a* 

the required volume is twice the volume 
generated by OAB^ we get, substituting 
in (A), 

2 •/© Jo a^^ 

2ira6a 



.-. F = 



3 
4ira63 




4 TO* 



To verify this result, let 6 = a. Then V = — — » the volume of a sphere, which 

is only a special case of the ellipsoid. When the ellipse is revolved about its major 
axis the solid generated is called a prolate spheroid ; when about its minor axis, an 
oblate spheroid. 



886 INTEGRAL CALCULUS 



1. Find the volume of the sphere generated by revolying the circle x^ + 1/^ = 1^ 
about a diameter. . ^"^y* 

3 

2. Find by integration the volume of the right cone generated by revolving 
the line joining the origin to the point (a, h) about the axis of X. yafe* 

3 

3. Find the volume of the cone generated by revolving the line of Ex. 2 about 

the axis of F. . iro«6 

Ans. • 

3 

4. Find the volume of the paraboloid of revolution generated by revolving the 
arc of the parabola y^ = 4 ox between the origin and the point (zi, ^i) about its axis. 

Aim, 2 waz^ = -^ — ; i.e. one half of the volume of the circumscribing cylinder. 

5. Show that the volumes generated by revolving y ^f about OX and OF are 
- and 2 w respectively. 

6. Find the volume generated by revolving the arc in Ex. 4 about the axis of F. 
Am. — ^ = \ vxi*yi ; i.e. one fifth of the cylinderof altitude yi and radius of base Xi, 

7. Find by integration the volume of the cone generated by revolving about 
OX that part of the line 4x — 5^ + 3 = which is intercepted between the coordi- 
nate axes. ^ 9 r 

Ans, — . 
100 

8. Find the volume of the spindle-shaped solid generated by revolving the 

hypocycloid x* + y* = a* about the axis of X, . 32 to* 

Ans. • 

105 

Scfi 

9. Find the volume generated by revolving the witch y = about its 

asymptote OX. x" + 4 a« 

Ans. 4TV. 

10. Find ther volume generated by revolving about OY that part of the parabola 

X* -f- y* = a* which is intercepted by the coordinate axis. wa^ 

Ans. — • 
15 

11. Find the volume of the torus (ring) generated by revolving the circle 
x2 + (y - 6)2 = a* about OX, Ans. 2 r^a^. 

12. Find the volume generated by revolving one arch of the sine curve y = sinz 

about the axis of X, H 

Atis. — -• 
2 

13. Find the volume generated by revolving about OX the curve 

(x — 4 a) y* = ax (x - 3 a) 
between the limits x = and x = 3 a. Ans. —(15 — 16 log 2). 



INTEGRATION A PROCESS OF SUMMATION 387 

14. Find the volume generated by revolving one arch of the cycloid 



« = r arc vers- — V2 ry — y* 

T 

about OX, its base. 



Hint. SubBtitute dx - ^ " , and limits y » 0, y » 2 r, in (A), p. 386. 

V2ry-y* Ana. 6tV*. 



X X 



15. Find the volume generated by revolving the catenary y = - (e^ + e «) about 
the axis of X from x = Otox = &. 

Ana. -g-(«--e °) + -y- 



16. Find the volume of the solid generated by revolving the cissoid ^ =: 



about its asymptote x = 2 a. 2 a « 

Ana, 2ir«a». 

17. Given the slope of tangent to the tractrix — = ^ ; find the solid 

generated by revolving it about OX ^ Vo^^ya 

Ana, ^waK 

18. Shovr that the volume of a conical cap of height a cut from the solid gen- 
erated by revolving the rectangular hyperbola x^ — y^ = a^ about OX equals the 
volume of a sphere of radius a. 

19. Using the parametric equations of the hypocycloid 

(X = a co8« By 
y = a sin« ^, 

find the volume of the solid generated by revolving it about OX. Ana. 

20. Find the volume generated by revolving one arch of the cycloid 

tx = a(^ — sin^), 
y= a(l -cos^), 
about its base OX. Ana. 6ir^*. 

Show that if the arch be revolved about OY the volume generated is 6 ir*a'. 

21. Show that the volume of the egg generated by revolving the curve 

xV = (X - a) (X - 6) 

about OX is » J (« + &)log~ - 2 (6 - a) | . 

22. Find the volume generated by revolving the curve x* - o%c« + aV = 

about OX. . 4 ira» 

Ana. • 

15 

23. Find the volume of the solid generated by revolving the curve x* 4- y' = 1 

about OF. . 4ir 

Ana. -— • 

5 



388 INTEGRAL CALCULUS 

228. Areas of surfaces of revolution. Let S denote the area of 

the shaded surface geuerated by revolving the arc CP{=«) about 

OX, the equation of 

the plane curve CPB 

being y=f(x). 

Let x(=OM) take 
on a small increment 
£^{=MN); then S 
takes on an increment 
AiS, the area of the band 
generated by the arc 
P(?(=A«). Draw the 
chord PQ. Let ^S' denote the area of the convex surface of the 
frustnim of the cone of revolution generated by the chord PC- 
Then 

A5' = ^^^ + ^^^y + ^y> -ebordPQ. 

rTbe Uitenl vm of the f milmm at ■ cone of reTolntion li equftl to one luin 
Lthe gum ot the clrc'unifereiiMa ot iti buea niulllplled bjr the lUnt height.] 



.(,.f). 



Multiplying and dividing the first member by AiS', and then 
dividing both members by A«, we get 



Now let Ax approach zero as a limit. Then At and Ay also 
approach zero and 

limit /A5\ _ dS limit /A.S'\ _ limit / chord PQ \ _ 
Ai^Oy^Bj rf/ Ai = 0\AsJ 'A« = 0^ A» )~ 

Hence from (A), 

d»~ 
or, using differentials, 

dS = 2 Trydt, 



INTEGRATION A PROCESS OF SUMMATION 889 

which is the differential of the area of the surface of revolution. 
Integrating, 

^ = 2 TT I yds ; 
or, since from (29), p. 143, ^* = 1 + ( "t^ ) ^^j we have 

2-1 4 



^=^'/^D-(f)> 



Hence if 0-4 = a and OB = 6, the surface generated by revolving 
the arc CD about OX is given by the formula 

w> «=="r«'[i+(g)']''^. 

6*2/ 

where the value of y and -y- in terms of x must be substituted 

dx 

from the equation of the given curve. 

This formula is easily remembered if we consider a narrow band 
of the surface included between two planes perpendicular to the 
axis of revolution as the element of area, and regard it as the 
convex surface of a truncated cone of revolution of infinitesimal 
slant height ds and with a middle section whose circumference 
equals 2 Try, hence of area 2 iryds. SuiAming up all such bands 
(elements) from A to By we get the area of the surface generated 
by revolving the arc CPD about OX. 

Similarly when OF is the axis of revolution we use the formula 

,c «=2,jrv[i+(g)-]W 

where the value of x and -^ in terms of y must be substituted 

dx 

from the equation of the given curve. 

Ex. 1. Find the area of the surface of revolution generated by revolving the 
hypocycloid x' 4- y' = a' about the axis of X. 

Solution. Here ^ = - ?^ , y = (a* - x')l 

dx ri 



190 INTEGRAL CALCULUS 

SubatiCating lu (B), p. S80, noting that the arc BA geoerates onlf one half of 
tha surface, we get 



= 2-j;"(a'-*')*(^)'dx 



_ 6Ttt' 

6 
12 *a* 



1. find the area of the mirface of the sphere generated b; revolving the circle 

!C« + j/« = H about a diameter. Ana. irt*. 

2. Find the area of the surface generated by revolving the parabola y* = 4 ox 
■bout OX, from the origin to the point where x — Za. a ^ i 

3. Find by iotegraUon the area of the surface of the cone generated by revolr- 
ing the line joining the origin Ui the point (a, b) about OX. ^^^ ,^ Va* + 6». 

4. Find the surface generated by revolving the catenarj y = - {^ +e ') about 
0¥ltomx = OUtx = a. ^ 

Am. 2wa'(l -«-'). 

5. Find the surface of the .prolate spheroid generated by revolving the ellipse 



V' = (1 - e^ (a' - X*) about OX, 



Am. 2 t63 4 - 



6. Find the surface of the torus (ring) generated by revolving the circle 
a!» + (V - 6)* = o> about OX. Aw. i w^ab. 

/Tinl. tJ>LDg (he positive value of Voi-i" gives the ontaide lorfaw, and the negatiTe vilne 
the lulde mrf ace. 

7. Find the surface generated by revolving an arch of tlie cycloid 

a: = rar^versJ-V2ry-i,' „^ 

about 1(8 base. Ant. 

3 

8. Find the surface generated by the cycloid when revolved about the tangent 
at iU vertex. . S2 vr» 

A^. -^. 

9. Find the surface of the oblate spheroid generated by revolving the ellipse 

nV + Wt* = o^* about its minor axis. , „ . 6» , 1 + e 

' Ant. 2ira' + ir — log-^!^. 

Hinl. (^eawDtrlclt^ofelllpBe. ' '~' 



INTEGRATION A PROCESS OF SUMMATION 891 

10. Find the surface generated when the cycloid is revolved about its axis. 

Ana. 8»f«(«--J). 

11. Find the surface generated by revolving about OX that portion of the curve 
y = e* which is to the left of the axis of F. ^j^g^ ■-['v^-f lo^(l + V2)]. 

12. A quadrant of a circle of radius a revolves about the tangent at one extrem ity ; 
prove that the area of the curved surface generated is ir (r — 2) aK 

13. Find the surface generated by revolving the cardioid 

X = a (2 COB $ — cos 2 0)^ 

» = a(28in»-8in2»), ^^^ 

about OX. Ans. — - — . 

6 

14. Show that the surface generated by revolving the cycloid 

ix = a(^-sin^), 
y = a(l — cos^), 

about OX is . 

3 

16. Show that the surface generated by revolving the curve x* — ah^ + 8 oV = 
about OX from x = Otox = ais-^ C"(Sa^ - 2x«)dx = -ira*. 

16. Show that the surface generated by revolving x* + 8 = 6 xy about OT from 
x = ltox = 2iBir( hlog2^j and that when revolved about OX it ^-rjr* 

17. Show that the surface generated by revolving the cubical parabola y = x* 
about OXfromx = 0tox = lis2ir f Vl + 9x*x«dx = ^ ( VlOOO - 1). 

la Show that if we rotate i/« + 4x = 2 logy about OX from y = 1 to y = 2, 
the surface generated is — — -. 



CHAPTER XXXI 

SUCCESSIVE AND PARTIAL INTEGRATION 

229. Successive Integration. Corresponding to succe89ive differ- 
entiation in the Differential Calculus we have the inverse process 
of ruceessive integration in the Integral Calculus. We shall illus- 
trate by means of examples the details of this process, and show 
how problems arise where it is necessary to apply it. 

Ex. 1. Given — ^ = 6x; to find y. 
Solution. We may write this 

4-) 

' = Ox, 



dx 

<i(g) = e«fa. 

Integrating, d^"^}^ ^^^' 



or, ^ = 3x* + ci. 



This may also be written 



dz' 



it) 



^dx „ . 

— = 3x« + ci, 

dx 

or, d(^) = (3x2 + ci)(fx. 

Integrating again, -^ = 1(3 x\-^ ci) dx, or, 

dx "^ 

(-4) J? = X8 + CiX + C2. 

dx 
Again, dy = (x' + CiX + Cj) dx, and integrating, 

/ «v X* CiX' 

(B) y = + 4. Cax + Ca. Arts. 

4 2 

The result (A) is also written in the form 

■892 



SUCCESSIVE AND PARTIAL INTEGRATION 393 

and is called a double integral^ while (J5) is -written in the form 
^ = I I I 6 xdxdxdx (or = I I I 6 xda?)^ 

and is called a triple integral. In general, a multiple integral 
requires two or more successive integrations. As before, if there 
are no limits assigned, as in the above example, the integral is 
indefinite; if there are limits assigned for each successive inte- 
gration, the integral is definite. 

Ex. 2. Find the equation of a curve for every point of which the second deriva- 
tive of the ordinate with respect to the abscissa equals 4. 

Solviion, Here — ^ = 4. Integrating as in Ex. 1, 

(D) y = 2 x^ + CiX 4 Cj. Ans. 

This is the equation of a parabola with its axis parallel to OY and extending 
upward. By giving the arbitrary constants of integration Cx and 0% all possible 
values we obtain all such parabolas. 

In order to determine Ci and Cs, two more conditions are necessary. Suppose 
we say (a) that at the point where x = 2 the slope of the tangent to the parabola is 
zero ; and (b) that the parabola passes through the point (2, — 1). 

(a) Substituting x = 2 and -^ = in (C) 

ox 

gives = 8 + Ci. 

Hence Ci = — 8, 

and (D) becomes y = 2 x^ — 8x + Ca. 

(b) The coordinates of (2, — 1) must satisfy this equation ; therefore 

— 1 = 8 — 16 + Ca, or, Ca = + 7. 
Therefore the equation of the particular parabola which satisfies all three condi- 
tions is y = 2xa-8x + 7. 

EXAMPLES 

d'l/ QX^ CiX^ 

1. Given -^ = ax« ; find y. Ans. y = -— + -7- 4- CaX + Cs. 

dx8 60 2 

2. Given — ^ = ; find y. Ana. y = — — + Cjx + Cs. 

dx' 2 

2 dx' CiX^ 

3. Given d«y = ; find y. Ana. y = logx + —- + CjX + cs- 

x" 2 

4. Given — = sin ^ ; find p. Ana. p = cos ^ + -|- + Ca^ + Cg. 

(W» 2 



894 INTEGRAL CALCULUS 

5. Given 3^ = ^^ - p J ^^^^' ^^' * = ^5 " o ^^^^ + ^ + ^«^ + ^* 

6. Given <Pp = sin^ co8^<pdiffl\ find p. Ans, p = ^lILt - _ gin + ci4> + cj. 

6 3 

7. Determine the equations of all curves having zero curvature. 

SinL ^=0, from(38),p. 161,8lno©ir=0. 
djfl 

Ana, y = cix^ cs, a doubly infinite system of straight lines. 

8. The acceleration of a moving point is constant and equal to /; find the dis- 
tance (space) traversed. 

Hint. ftomKx.28.p.lM,^'-/. AnS. 8 = ^ + C^t + C 

9. Show in Ex. 8 that Ci stands for the initial velocity and Ct for the initial 
distance. 

10. Find the equation of the curve at each point of which the second derivative 
of the ordinate with respect to the abscissa is four times the abscissa, and which 
passes through the origin and the point (2, 4). Ans. 3y = 2z(2* — 1). 

d*y Cix' Mt^ 

11. Given —^ = xcosx; findy. -4»w. y=xcosx— 4sinx + ---H hc»x + c«- 

(2x* 6 2 

«A r,, d'y , . « , . 7cosx cos*x . CiX* . 

12. Given -~ = sin«x ; findy. Ana, y = — — - + -^ +C2r +C8. 

(IX V «7 Jt 

230. Partial Integration. Corresponding to partial differentior 
lion in the Differential Calculus we have the inverse process of 
partial integration in the Integral Calculus. . As may be inferred 
from the connection, partial integration means that, having given 
a differential expression involving two or more independent vari- 
ables, we integrate it, considering first a single one only as varying 
and all the rest constant. Then we integrate the result, considering 
another one as varying and the others constant, and so on. Such 
integrals are called double^ triple^ etc., according to the number of 
variables, and are called multiple integrals.* 

Thus the expression 

u=j jf{x, y)dydx 

indicates that we wish to find a function t^ of 2; and y such that 

* The integrals of the same name in the last section are siMCial cases of these, namely, when 
we integrate with respect to the same variable throughout. 



SUCCESSIVE AND PARTIAL INTEGRATION 395 

In the solution of this problem the only new feature is that 
the constant of integration has a new form. We shall illustrate 
this by means of examples. Thus, suppose we wish to find u^ 
having given 

P = 2a: + y + 3. 

OX 

Integrating this with respect to x, considering y as constant, 
we have ^ o j. 

where ^ denotes the constant of integration. But since y was 
regarded as constant during this integration, it may happen that <f> 
involves y in some way ; in fact ^ will in general be a function of y. 
We shall then indicate this dependence of ^ on y by replacing 
<f> by the symbol ^(y). Hence the most general form of u is 

where ^(y) denotes an arbitrary function of if. 
As another problem let us find 

(A) u=ff{2^ + }/')dydx. 

This means that we wish to find u, having given 

— — ^-\-y* 

dxdy 

Integrating first with respect to y, regarding x as constant, 
we get . 

where '^(2:) is an arbitrary function of x and is to be regarded as 
the constant of integration. 

Now integrating this result with respect to a;, regarding y as 
constant, we have 

where <E> (y ) is the constant of integration, and 

^(2:)= Cylr{x)dx. 



396 



INTEGRAL CALCULUS 



231. Definite double integral. Geometric interpretation. Let 
/(^i y) be a continuous and single-valued function of x and y. 
Geometrically, 

{A) z =f{x, y) 

is the equation of a surface, as KL. Take some area S in the XY 
plane and construct upon iS^ as a base, the right cylinder whose 
elements are accordingly parallel to OZ, Let this cylinder inter- 
sect KL in the area ^S^, and now let us find the volume V of the 
solid bounded by S^ S\ and the cylindrical surface. We proceed 
as follows. 

At equal distances apart (= Aa;) in the area S draw a set of lines 
parallel to OF, and then a second set parallel to OX at equal 




distances apart (= Ay). Through these lines pass planes parallel 
to YOZ and XOZ respectively. Then within the areas S and S 
we have a network of lines, as in the figure, that in S being 
composed of rectangles, each of area Ax-^y. This construction 
divides the cylinder into a number of vertical columns, such as 
MNFQj whose upper and lower bases are corresponding portions 
of the networks in S* and S respectively. As the upper bases of 
these columns are curvilinear, we of course cannot calculate the 
volume of the columns directly. Let us replace these columns 
by prisms whose upper bases are found thus : each column is cut 
through by a plane parallel to XY passed through that vertex 



SUCCESSIVE AND PARTIAL INTEGRATION 897 

of the upper base for which x and y have the least numerical 
values. Thus the column MNPQ is replaced by the right prism 
MNPR^ the upper base being in a plane through P parallel to 
the XOY plane. 

If the coordinates of P are (a:, y, z), then MP = z =/(a:, y), and 
therefore 

{B) Volume of MNPR =/(x, y) Ay -iix. 

Calculating the volume of each of the other prisms formed in 
the same way by replacing x and y in (-B) by corresponding values, 
and adding the results, we obtain a volume V* approximately equal 
to F; that is, 

((7) r=22/(^y)^y-^^5 

where the double summation sign XX ^^^i^*^ ^^* there are two 
variables in the quantity to be summed up. 

If now in the figure we increase the number of divisions of 
the network in S indefinitely by letting Aa: and Ay diminish 
indefinitely, and calculate in each case the double sum ((7), then 
obviously F' will approach F as a limit, and hence we have the 
fundamental result 

limit » . 

(D) F=Ay = oXX/(^' y)Ay.A2;. 

Let us see how to calculate this double limit. We commence 
by calculating ((7) for all the prisms of a row parallel to YOZ^ q?lj 
along the line DG. 

This will give us, approximately of course, the volume of a 
slice of F bounded by planes through P and Q parallel to the 
YOZ plane. To do this analytically, we sum up in ((7), keeping 
X constant (= 02>). Since in this summation Ax is also constant, 
we may write {C) in the form 

(^) F' = XA2:.X/(2r,y)Ay. 

Hence (2>) becomes 

limit . .. 

(F) r= A» = y Ax . y /(x, y) Ay. 



398 INTEGRAL CALCULUS 

In (J?), the limits for the second sign of summation are the 
extreme values of y for the vertices of the network along the 
line i>&, and for the first sign of summation the extreme values 
of 2; in the entire network. Hence it should now be intuitianaUy 
evident that (F) becomes* 



dxi f{x,y)dy, 

HA ^ np 



for we have merely replaced the signs of summation by integral 
signs, and the limits by the values taken from the region S itself. 
We have accordingly the fundamental result, 

limit _^^^ r^^ r^<^ 

(E) F=Ay = oVV/(z,y)Ay.Aa:= j j f{x,y)dydx, 

the second integration sign applying to y and being performed 
first, X and dx being meanwhile regarded as constants. 

The process of evaluating (2>) might have been carried out by 
first adding up the columns in ((7) along a line parallel to OX, 
i.e. y remaining constant, afterwards summing up the resulting 
prisms by varying y, and finally passing to the limit as Aa: and Ay 
approach zero. These steps would be indicated by writing the 
differential expression in (F) in the form 

/(a:, y)dxdy 

and changing the limits. Summing up our line of reasoning, we 
may write 

limit . . /•«! /••*! 

F=Ay = oVy/(ar,y)Ay.Ax= j I f{x,y)dydx 






f{x, y) dxdy, 



where v^ and v, are in general functions of y, and u^ and u^ func- 
tions of a:, the second integral sign applying to the first differential 
and being calculated first. 

•A rigorooB proof of (G) is to be found In Goursat's Cour$ d*Analy»e Mathimatique, 
Vol. J, §123. An English translation, by Professor Uedrick, of this book is published by 
Giun & Company. 



SUCCESSIVE AND PARTIAL INTEGRATION 

Our result may be stated in the following fotm : 
The definite double integral 



Crf(^^y)dydx 



may he interpreted at that portion of the volume of a truncated 
right cylinder which it included between the plane XOY and the 

■> z =f{x, y), 

the bate of the cylinder being the area bounded hy the curvet 
y =.Uy y=u^ x=a^, x=. u,. 

Similarly for the second integral. 

It is instructive to look upon the above process of finding the 
voliime of the solid as follows: 

Consider a column of infinitesimal base dydx and altitude z as 
an element of the volume. Summing up all such elements from 
y = DF to y = DG, x in the meanwhile being constant (say = OD), 
gives the volume of an indefinitely thin slice having FGJil as one 
face. The volume of the whole solid is then found by summing 
up all such slices from x = OA to a; = OB. 

In partial integration involving two variables the order of inte- 
gration denotes that the limits on the inside integral sign correspond 
to the variable whose differential is written inside, the differentials 
of the variables and their corresponding limits on the integral signs 
being written in the reverse order. 

Ex. 1. Find the volae of the definite double iut«gnl 
J^-^ "''"{x + v)dvdx. 

Solution. J C it+y)dydx 



400 INTEGRAL CALCULUS 

Interpreting thin result geometHcally it means that we have found the Tolume 
of the solid of cylindrical shape ataudlng on OAB aa base and bounded at the top 
by the aurfacs (plane) z — z -k-y. 

The attention of the student is now particularly called to the maimer in which 
the limits do bound the base OAB, which coirespondB to the area S in the figure 
p. 306. Our solid here stands on a baae in the XY plane bounded by 

V = (line OB) i 

r I from y limits : 

V = va" - 3fl (quadrant of circle AB) ) 

» = (line OA) 1 
x = a (line BE) \ 

232. Value of a definite double integral over a region S'. In the 
last section we represented the definite double integral as a 
volume. This does not necessarily mean that every definite 
double integral is a volume, for the physical interpretation of 
the result depends on the nature of the quantities represented by 
x^ y, z. Thus, if Xy y, z are simply considered as the coordinates 
of a point in space, and nottting more, then the result is indeed a 
volume. In order to give the definite double integral in question 
an interpretation not necessarily involving the geometrical concept 
of volume, we observe at once that the variable z does not occur 
explicitly in the integral, and therefore we may confine ourselves 
to the XY plane. In fact, let us consider simply a region S in the 
XY plane, and a given function /(r, y). Then, drawing a network 
aa before, calculate the value of 

for each point of the network, and sum up, finding in this way 

and finally pass to the limit as Ax and A^ approach zero. This 
operation we call integrating the function f{x, y) over the region S, 
and it is denoted by the symbol 



}jfi?^y)dydx. 
S 
>ounded by the cu: 
x=aj, ■y = u^, y. 



If S is bounded by the curves 
x = a^, x= a,, y = it,, y = u^, then 



SUCCESSIVE AND PARTIAL INTEGRATION 401 

We may state our result as follows : 

• 

To integrate a given function f{x^ y) over a given region S in the 
XOY plane means to calcvlate the value of 

limit ^ ^ 

Ax = yy.f(x, y) AyAx, 
Ay = ^ ^ 

as explained ahove^ and the result is equal to the definite double 
integral 

r r /(^ y) dydoi, or f C^f{x, y) dxdy, 

the limits being chosen so that the entire region S is covered. This 
process is indicated briefly by 



fjfi^^ y) ^y^^' 



In the sections which follow we shall show by numerous exam- 
ples how the area of the region itself may be calculated in this 
way, and also the moment of inertia of the region. 

Before attempting to apply partial integration to practical prob- 
lems it is best that the student should acquire by practice some ' 
facility in evaluating definite multiple integrals. 

r (a-y)xadydx = -— . 

Jr»8* /•" /•s^r 2/*l" r^a^ 7a26* 



I xdydx = — — 



Solution. I I xdydx = ( \xy\ dz 



=j(;"2xv;^3^<fa=[-|(a.-«.y];=?«' 



In partial integration involving three variables the order of 
integration is denoted in the same way as for two variables ; that 
is, the order of the limits on the integral signs, reading from the 
inside to the left, is the same as the order of the corresponding 
variables whose differentials are read from the inside to the right. 



402 INTEGRAL CALCULUS 



•i /.I z*] 



Ex. a. Verify f f f xt/'dzdydx = ^. 

Solution, c f c "^^'^^yot" = c r r X *!'**']<'!"'* ~ jc f r*""*! '**<** 



85 
2 



EXAMPLES 

Verify the following. 

2 

5. r f r^ain ^d^r = — ^ — (cos/S — cos a). 
Jo 3 

ph ftVSt . 

12. JT J Vrt-«ad«a = 6&». 



Ja Jv ^ ' 30 



233. Plane area as a definite double integral. Rectangular coordi- 
nates. As a simple application of the theorem of the last section 
(p. 401), we shall now determine the area of the region S itself in 
the XO Y plane by double integration.* 

* Some of the ezamplefl that will be given in this and the following articles may be solved by 
means of a single integration by methods already explained (§§^1« ^^). The only reason in 
such cases for using successive integration is to familiarize the student vith a new method for 
solution which is sometimes the only one possible. 



SUCCESSIVE AND PARTIAL INTEGRATION 



403 



As before, draw lines parallel to OX and F at distances Lx 
and Ay respectively. Now take 

element of area = area of rectangle FQz=z£ki/- Ax^ 

the coordinates of P being (a;, y). 

Denoting by A the entire area of region Sj we have 

limit .^. 

(A) ^ = Ax = oVVAy.Aa:. 

We calculate this by the theorem 
on p. 401, setting /(ar, y) = l, and get 

OB /%C£ 



I dydx, 

OA %/ r.n 




where CD and CE are in general 
functions of x^ and OA and OB are constants giving the extreme 
values of x^ all four of these quantities being determined from the 
equations of the curve or curves which bound the region S. 

It is instructive to inteipret this double integral geometrically 
by referring to our figure. When we integrate first with respect 
to y, keeping x (=0C) constant, we are summing up all the 
elements in a vertical strip (as DF). Then integrating the result 
with respect to x means that we are summing up all such vertical 
strips included in the region, and this obviously gives the entire 
area of the region S, 

Or, if we change the order of integration, we have 



(0) 



OL /%m 



I dxdy^ 

QIC ^na 



where HG and HI are in general functions of y, and OK and OL 
are constants giving the extreme values of y, all four of these quan- 
tities being determined from the equa- 
tions of the curve or curves which 
bound the region S. Geometrically 
this means that we now commence 
by summing up all the elements in 
a horizontal strip (as G«7), and then 
find the entire area by summing up 
all such strips within the region. 



• 


r 




»— 1 












^ 


< 


^ 

-k^— 




_\ 




. 


^l*^ 










z 


1 




_j 




-- o|^ 




Q 














Jt 


vtr 


b'i 


I 


f^'' 


' 




' .■ 


,-/ ■ 


"k 


f*J 


H 


pjl 


m 
t 










J 


Vi 


« 




\ 














> 


Y 




K 


•- — —• — • 


^ 


•>* 






^ 


Ix^ 


v 







c 


y 
f 



















404 



INTEGRAL CALCULUS 



Corresponding to the two orders of summation (integration), the 
following notation and figures are sometimes used : 



(i>) 



A= j j dydx ; A= i i dxdy. 

S S 



o 




o 




Referring to the result stated on page 401, we may say: 

The area of any region is the value of the dovhle integral of the 
function f{x^ y) = 1 taken over that region. 

Or, also, from § 231, p. 399, 

The area equals numerically the volume of a right cylinder of unit 
height erected on the base S. 

Ex. 1. Calculate the area of the circle x* + y* = r* by double integration. 
SolvtUm. Summing up first the elements in a vertical strip, we have from 

(B), p. 408, pOApJiR 

^ = I I dydz. 
Job Jus 

From the equation of the boundary curve (circle) 

.X 0B = - r, OA = r. 

I ^ dydz • 

■ r J-V^^^^ 




Hence 



A = 



- 2 C'^^r'^ - x»dx = irt^. Am. 



When the region whose area we wish to find is symmetrical 
with respect to one or both of the coordinate axes, it sometimes 
saves us labor to calculate the area of only a part at first. In the 
above example we may choose our limits so as to cover only one 
quadrant of the circle, and then multiply the result by 4. Thus, 

— = I I dydx = I Vr^ — ^dx = 

.*. A = Trr". Ans. 



irr 

T 



SUCCESSIVE AND PARTIAL IKTEGRATION 



405 



Ex. 2. Calculate that portion of the area which lies above OX bounded by 
the semicubical parabola ^ = x" and the straight line 
y = x. 

Solution, Summing up first the elemenU in a hori- 
zontal strip, we have from (C), p. 403, 

o Jab 

From the equation of the line, ^B = y, and from the 
equation of the curve, AC = y^, solving each one for x. 
To detenu ine OD^ solve the two equations simultane- 
ously to find the point of intersection E, This gives the 
point (1, 1) ; hence OD = 1. Therefore 

''=rjr'"'=r<^-«*=[¥-?]: 




1. Find the area of the ellipse — f- ^ = 1 by double integration. 

«» fta 

Ans, 4 11 dydx = ir od. 

2. Find by double integration the area between the straight line and a parabola 
with its axis along OX, each of which joins the origin and the point (a, b), 

I "dydx = — . 

•/to 6 

a 

3. Find by double integration the area of the rectangle formed by the coordi- 
nate axes and two lines through (a, b) parallel to the coordinate axes. Ans. a6. 



4. Find by double integration the area of the triangle formed by OJC and the 



lines X = a and y = -x. 

a 



, ab 
Ans. — . 
2 



6. Find by double integration the area between the two parabolas 3 ^ = 25 x 
and6x« = 9y. Ans. 5. 



6. Required the area in the first quadrant which lies between the parabola 



f/^ = ax and the circle y* = 2 ax — x^. 



An8, 



IT a» 2 a2 



7. Find the area outside of the parabola y^ = 4 a (a — x) and inside of the circle 
y« = 4 aa - x2. 



406 INTEGEAL CALCULUS 

8^ Solve tha problenu on page 406 by flrat aumming up all the elementB in a 
horizontal BUip, ftnd theo •ummlDg up all each atripe. 

^n*. Ex. 3, C C° dxdy = ab. Ei. 4, f C^dxd» = ~. 

234. Plane area as a definite double integral. Polar coordinates. 
Suppose the equations of the curve or curves which bound the 
region S are given in polar coordinates. Then the region may 
be divided into checks bounded by 
mdial lines drawn from the origin, 
each one making the angle A0 with 
the next one, and concentric circles 
drawn with centers at the origin, the 
difference between each radius and 
the next one being Ap. 

We shall consider these checks as 
the elements of area of the region S. Let us calculate the area 
of one, say FQ, bounded by arcs with radii p and p + Af>. 
From Geometry, 

area of sector OSQ = ^ (/> + Ap)*£i0, 
area of sector OPS = J f^A0. 
Hence 

area of element FQ = ^(j> + A^)' Al? — ^ p^M 
= pA/)A5 + ^ Ap" ■ Ad. 
Then, aa before, the area of the region S will be 

limit _,_. , 

^=4j = 05;^(pApA6' + iA?Afl) 

=ffpdpd9 + f^=li^P-hXX^P^0, 
the summation extending over the entire region, or, 
{A) A =j'j'pdpd$. 

S 
Here, again, the summation (integration) may be effected in 
two ways. 



SUCCESSIVE AND PAETIAL INTEGHATION 407 

When we integrate first with respect to &, keeping p constant, 
it means that we sum up all the elements (checks) in a segment of 
a circular ring (as ABCD), and next integrating with respect to p. 
that we sum up all such rii^ within the entire region. Our 
limits then appear as follows : 



<^ ^-£'XI 



the angles XOA and XOB being in general fimctions of />, and 
OE and OF constants giving the 
extreme values of p. 

Suppose we now reverse the 
order of integration. Integrating 
first with respect to /», keeping 
constant, means that we sum up 
all the elements (checks) in a 
wedge-shaped strip (as GKLH). 

Then integrating with respect to 6, we sum up all such strips 
within the region S. Here 



XmntitXOI f%Oa 
I pdpdd, 
_wi. xoj 'foo 



OH and OG being in general functions of B, and the angles XOJ 
and .^OJ being constants giving the extreme values of 6. 

Corresponding to the two orders of summation (integration), the 
following notation and figures may be conveniently employed : 



(i)) . A =j'j'pd0dp, A =j'j'pdpde. 




These are easily remembered if we think of the elements (checks) 
as being rectangles with dimensions pdd and dp, and hence of area 
pdddp. 



408 



INTEGRAL CALCULUS 



Ex. 1. Find the area of the circle p = 2r cob $ by double integration. 

Solution, Summing up all the elements in a 
sector (as 0£), the limits are and 2r cos ^; and 
summing up all such sectorSi the limits are and 

- for the semicircle OXB. Substituting in (D), 




A 



2 Jo Jo 

A = irf*. AtO, 



2reot9 ,ry4 

pdpde = — , or, 



EXAMPLES 

1. In the above example find the area by integrating first with respect to $, 

2. Find by double integration the entire area of the cardioid p = a (1 — cos S), 

3ro« 
Ana. • 



3. Find by double integration the entire area of the lemniscate p^^a^ cos 2 $, 

Am. a^ 

4. Find by double integration the area of that part of the parabola ^ = a sec' - 
intercepted between the curve and its latus rectum. 

/•- /•atec"- QaS 



^- ^XU" ^'^^= 



3 



6. Find by double integration the area between the two circles ^ = a cos B^ 
p = b cos $fb>a'f integrating first with respect to p, 

6. Solve the last problem by first integrating with respect to fi. 

235. Moment of inertia. Rectangular coordinates. Consider an 
element of the area of region S^ as FQ, 
If the coordinates of P are (r, y), the 
distance of P from is Var^ 4- 1/\ Mul- 
tiplying the area of element, i.e. 

ArAy, 

by the square of the distance of P from 
the origin, we have 

(A) (i^^f/^AyAx. 




SUCCESSIVE AND PARTIAL INTEGRATION 



409 



Then the value of 



limit 



6x = Q^^ 



defines the moment of inertia of the area within the region S about 
the point 0, when the summation is extended over the entire region. 

The product (A) is then an element of the moment of inertia. 

Denoting this moment of inertia by /, we then have, as before, 



(C) 



J= C C(x* + y^)dydx» 



8 



the limits of integration being determined in the same way as for 
finding the area (p. 403). 

Ex. 1. Find I over the area bounded by the lines x = a, y = 0, y = -x. 

Solution. These lines bound a triangle OAB. Summing up all the elements 
in a vertical strip (as PQ), the y-limits are zero and - x 



a 
(found from the equation of the line OB). Summing 

up all such strips within the region (triangle), the 

x-limits are zero* and a(= OA). Hence 

a2 b* 



B(aP) 




= ab( — I ) . Ans. 

\4 12/ 

If we suppose the triangle to be composed of horizontal strips (as RS)^ 



EXAMPLES 

1. Find I over the rectangle bounded by the lines z = a, y = h^ and the 
coordinate axes. . ' ^T^ a . o\j ^ a96 + a6» 

2. Find I over the right triangle formed by the coordinate axes and the line 
joining the pomts (a, 0), (0, 6). h(a-x) 



3. Find I for the region within the circle z^ :f- y* = r*. 



Ans. 



irr* 



• From the reflult, p. 401, we may say that tJie mommt of inertia of the area feithin the region 
S is the value of the double integral of the function f (x, y)=x* + p* taken over that region. 



410 INTEGRAL CALCULUS 

4. Find I for the ellipse -^ + ^ =" 1- Compare reaolt nltli preceding probli 



Ant. '^{tfl-^-t^y 



5. Find J OTer Ihe region between the Btraight line and a parabola with axis 
along OX, each of which joins the origin and the point (a, b). 

6. Find I over the region bounded bj the paratwla y* = 4 oz, the line z + y 
- 6 a = 0, and OX. __ 

""•■ x'i'"" " + '^"•^ + j;"x""'<"" + rt"''^ = TT' 

236. Homeat of inertia. Polar coerdinates. Consider an ele- 
ment of the area of region S, as FQ. If the codrdinates of P 
are (p, 8), the distance of P from is p. The area of the element 

in polar coordinates was found, od 

p. 406, to be 

Multiplying this by the square of 
the distance of P from the otigin, 
** we get 

{A) p'ifiApAe+i^Ap.^e). 

Then, in conformity with the definition of moment of inertia 
(=/) of the last section, we say that 

(B) ir=o'^^p'(pApA0 + lAp.A0) 

dejinet the moment of inertia about of the area within the region S. 
Or, passing to the limit (as under § 234, p. 406), 



I=jjf^d$dp, 



the limits of integration being determined in the same way as in 
finding the area (p. 407). 



SL'CCESSIVE AND PARTIAL INTEGRATION 4 

Ex. 1. Find / over the region bounded by the circle p = 2rcoe0. 

SoliUioTu Summing up the elements in the triiuigalar-eliaped strip OP, i 
(>-limiiH are zero and 2 r cob 9 (found from the equation of the circle). 

Summing up all Buch Btrips, the 0-limits 
are and - . Hence 



'-fli"'"' ''*"'- 

amiag apt 
ip(u,QB), 



Summing up first the elementB in a ciicu- 
lar atrip (as QR), we have 




,. fi£-"-^^ 



its latuB Tectum, 
480* 



2. Find I over the entire area of the cardioid ^ = a(l — cos e). 

S. ««/«"=—• 

8. Find I over the area of the lemntocate p* = a> cos 2 9. , wO* 

An.. — . 

4. Find I oTer the area bounded hj one loop of the curve p = acoB2B. 

237. Geaeral method for 
finding the areas of sur- 
faces. The method given 
in § 228 for finding the area 
of a eurfnce applied only 
to surfaces of revolution. 
We shall now give a more 
general method. I*et 

(A) »=/(!, y) 
be the equation of the 
surface KL in figure, and 
suppose it is required to 
calculate the area of the re^on S' lying on the surface. 



412 INTEGRAL CALCULUS 

Denote by S the region on the JTOF plane, which is the orthogo- 
nal projection of S' on that plane. Now pass planes parallel to 
YOZ and XOZ at common distances Ax and Ay respectively. As 
in § 231, these places form truncated prisms (as FB) bounded at 
the top by a portion (as PQ) of the given surface whose projection 
on .the XOY plane is a rectangle of area /ixAy (as AB)y which 
rectangle also forms the lower base of the prism, the coordinates 
of P being {qc^ y, z). 

Now consider the plane tangent to the surface KL at P. Evi- 
dently the same rectangle AB is the projection on the XOY plane 
of that portion of the tai^gent plane (PB) which is intercepted by 
the prism PB, Assuming 7 as the angle the tangent plane makes 
with the XOY plane, we have 

area AB = area PB • cos 7, 

rThe projection of a plane area upon a second plane is equal to the area of thel 
[portion projected multiplied by the cosine of the angle between the planes. J 



or, AyAx = area PB . cos 7. 

1 



But cos 7 = 



[-(sy-d)'] 



»' 



rCosine of angle between tuigent plane, (70) , p. 274, and XO Fl 
Lplane found by method given in Solid Analytic Geometry.] 

area PB 



hence Ai/Ax = 



[-(DHDT' 



or. 



^.P.=[..(|)V(J;)-]W 



which we take as the element of area of the region /?. We then 
define the area of the region S' as 

the summation extending over the region ^, as in § 231. Denot- 
ing by A the area of the region S\ we have 

'« ^=//['+(i)*+(S)']'""-- 

S 



SUCCESSIVE AND PARTIAL INTEGRATION 418 

the limits of integration depending on the prafection on the XOY 
plane of the region whose area we wish to calculate. Thus for (B) 
we chose our limits from the boundary curve or curves of the 
region S in the XOF plane precisely as we have been doing in the 
previous four sections. 

If it is more convenient to project the required area on the XOZ 
plane, use the formula 

"^ ^=//[>+©'+(i)']*'^ 

8 

where the limits are found from the boundary of the region Sy which 
is now the projection of the .required area on the XOZ plane. 
Similarly we may use 

^=//['+(g)"+(S)l''-^- 

8 

the limits being found by projecting the required area on the YOZ 
plane. 

In some problems it is required to find the area of a portion 
of one surface intercepted by a second surface. In such cases 
the partial derivatives required for substitution in the formula 
should be found from the equation of the surface whose partial 
area is wanted. 

Since the limits are found by projecting the required area on 
one of the coordinate planes, it should be remembered that — 

To find the projection of the area required on the XOY plane^ 
eliminate z between the equations of the surfaces whose intersections 
form the boundary of the area. 

Similarly we eliminate y to find the projection on the XOZ plane^ 
and X to find it on the YOZ plane. 

This area of a surface gives a further illustration of integration 
of a function over a given area. Thus in (-B), p. 412, we integrate 
the function „ „ , 

over the projection on the XO Y plane of the required curvilinear 
surface. 



414 



ISTEGRAL CALCULUS 



Ex. 1. Find the ares 

iDtegiutlou. 

Solution. Let ABC u 



of the Hurface of sphere z* + v* + ** = t* by donMa 
the figure be one eighth of the surface o( the sphere. 



■-©'-(g)'= 



, y»_j' + y* + i'_ 



-t^-V* 



The projection of the area required on the JOy plane iaAOB, a region bounded 
by z = 0, (OB) ; y = 0, (OA) ; i" + ]/> = r", (BA). 

Integrating first with respect to y, ne sum 
up all the elements along a strip (as DEFG) 
which is projected on the XOY plane in a 
strip also (as MNFG); that is, our i/-limits 
are zero and AfP'(=Vr> -- 1"). Then integrat- 
ing with respect to x sums up all such strips 
composing the surface ABC; that is, our 
z-llmits at« zero and OA (= r). Snbeiltiitlng 
In (-*). 

J _ /■■■ n^r*-!* rdydx _ Tf* 
8 "J, Jo 



= 4«4. Am. 



Vj^ - js - 



The ct 






f whose b 



jre of radius r is on the surface of a right cylinder, 
Find the surface of the cylinder intercepted b; 



Ex.2. 

the radius o 

the sphere. 

Solution. Taking the origin at the center of the sphere, an element of the 
cylinder for the z-axis, and a diameter of a right section of the cylinder for the 
z-axis, the equation of the sphere is x' + v^ + z* =i i^, and of the cylinder 
x' + y^ = TX. ODAPB is evidently one fourth 
of the cylindrical surface required. Since this 
area projects into the semicircular arc ODA 
on the SOT plane, there is no region 5 from 
which to determine our limits in this plane ; 
hence we will project our area on, say, the 
XOZ plane. Then the region S over which 
we integrate is OACB, which is bounded by 
1 = 0, {OA); x = 0,{OB); i' + TX = T^,(AClf); 
the last equation being found by eliminating 
y between the equations of the two surfaces. 
Integrating first with respect to z means that 
we sum up all the elements in a verti cal strip 
(as PD), the i-limits being zero and Vr' — rx. 
Then on integrating witii respect to z we sum up ail such stripe, the x-limits being 

Since the required surface lies on the cylinder, the partial derivatives required 
for formula (C), p. 413, must be found from the equation of the cylinder. 




SUCCESSIVE AND PARTIAL INTEGRATION 415 

Hence — = , -i^ = 0. 

dz 2y dz 

Substitating in (C), p. 413, 

f=rj[~^['-('-iF^)T"- 

Substituting the value of y in terms of x from the equation of the cylinder, 
Jo Jo voic-a^ Jo Vrx-x^ Jo ^* 



EXAMPLES 

1. In the preceding example find the surface of the sphere intercepted by the 
cylinder. ^r ^Vrx-x^ dydx 

2. The axes of two equal right circular cylinders, r being the radius of their 
bases, intersect at right angles ; find the surface of one intercepted by the other. 
Hint. Take jfl + z*'=r* and a:* + y«= r« as equations of oylinders. 



Jo «w i/-.a _ /1.2 



Vra-x2 

3. Find the area of the portion of the surface of the sphere x^ -hy^ •{■ z^ = 2ry 

lying within the paraboloid y = ax^ -{- hz\ , 2 ttt 

Ans, —z^- 

4. Find the surface of the cylinder x^ + y* = f* included between the plane 
2 = mx and the XOY plane. Ans. 4 r^. 

6. Find the surface of the cylinder z^ + (x cos a + y sin o)» = r^ which is situ- 
ated in the positive compartment of co5rdinates. 

IRift/. The axis of this cylinder Is the line 2» 0, xoofla+y8ina*0; and the radius of base is r. 



Ans, 



sm a cos a 



6. The diameter of a sphere whose radius is r is the axis of a right prism with 
square base of side 2 a. Find the surface of the sphere intercepted by the prism. 



Ans. 8 r ( 2 a arc sin — — r arc sin — - — ) 



7. Find the surface of the sphere x^ + y* + a" = a^ in the first octant inter- 
cepted between the planes x = 0, y = 0, x = 6, y = &. 

Atvs. a ( 2 6 arc sin — — ^^=; — a arc sin |. 

238. Volumes found by triple integration. In many cases the 
volume of a solid bounded by surfaces whose equations are given 
may be calculated by means of three successive integrations, the 



416 INTEGRAL CALCULUS 

process being merely an extension of the methods employed in 
the preceding sections of this chapter. 

Suppose the solid in question be diirided by planes parallel to 
the coordinate planes into rectang^ar parallelopipeds having the 
dimensions A2;, Ay, A2;. The volume of one of these parallelo- 

pipeds i8 Az.Ay.Ax, 

and we choose it as the element of volume. 

Now sum up all such elements within the region B bounded 
by the given surfaces by first summing up all the elements in a 
column parallel to one of the coordinate axes; then sum up all 
such columns in a slice parallel to one of the coordinate planes 
containing that axis, and finally sum up all such slices within the 
region in question. The volume V of the solid will then be the 
limit of this triple sum as A2, Ay, Ax each approaches zero as a 
limit. That is, 

limit 

the summations being extended over the entire region B bounded 
by the given surfaces. Or, what amounts to the same thing, 

F= j j j dzdj/doD, 
It 

the limits of integration depending on the equations of the bound- 
ing surfaces. 

Thus, by extension of the principle of § 232, p. 401, we speak 
of volume as the result of integrating the function f{Xy y, z) = 1 
throughout a given region. More generally many problems require 
the integration of a variable function of 2:, y, and z throughout a 
given region, this being expressed by the notation 

ffjA^^ y, z) dzdydx, 
R 

which is of course the limit of a triple sum analogous to the double 
sums we have already discussed. The method of evaluating this 
triple integral is precisely analogous to that already explained for 
double integrals in § 232, p. 401. 



SUCCESSIVE AND PARTIAL INTEGRATION 
Ex. 1. Find the volume ol th&t portion ol ihe ellipsoid 



wbich lies in the Bret octant. 

Sotution. Let O- ABC he that portion of 
the ellipsoid nhose volume la required, the 
equations ot the bounding surfaces being 

3!* tfl f 
<I) - + g + - = ,,(XBO. 

(2) * = 0, (OAB). 

W y = 0. (OACi, 

(4) 1 = 0, (OBC). 

PQ is an element, being one of the rectan- 
gular parallciopipede with dimensions Az, Ay, ix into which the planes parallel to 
the coordinate planes have divided the region. 

Inlegraling first with respect to j/r "« si™ "P *!' suc h elements in a colamn 
<aB flS), the *-limits being zero [from (2)] and TB = cyjl --,-^ [from (1) by 
solving for z]. " 

Iiitegrating next with respect to y, we sum np all such colum nH in a tiice 

(as DEMXGF), the tf-liinits being zero [from (3)1 and MQ = b-J\ - — [from 
equation of the curve AGB, namely ~x + ^ = ti ^y solving for y]. 

Lastly, Integrating with respect to z, we sum up all such slices within the entire 
region — ABC, the x-limit« being zero [from (4)] and OA = a. 

Hence ^~ i, f ""£ "" ^'^V^ 



-ilX""" 



"■"^ = Tr- 



Therefore the volume of the entire ellipsoid is 



Ex. 2. Find the volume of the solid contained 
between the paraboloid of 
revolution a:' + j/^ = az, 

the cylinder i" + v* = 2 ai, 

and the plane i = 0. 

Solution. The ^-limits are zero and NP{ = 
found by solving equation of parabolo id for z). " 

The i/-limit8 are zero and JlfJV(= Va m - *', found 
by solving equation of cylinder for y). 

Tbez-limitaare£eroandO.^(=2a). 



t' + y* 



418 INTEGRAL CALCULUS 

The above limitB are for the solid ONAB, one half of the solid whose yolmne is 
required. 



Hence 



Therefore V= 



EXAHPLBS 

1. Find the Yolome of the sphere z^ + ^^ + 2^ = v^ hy triple integration. 

Ajis, . 

3 

2. Find the volume of one of the wedges cut from the cylinder z^ + y* = v^ by 

the planes « = and z = mx, -r >*v^rt-a« ^mx 2 rhn 

Ana. 2 I I | dzdydz = . 

Jo Jo Jo 3 

3. Find the volume of a right elliptic cylinder whose axis coincides with the 
z-azis and whose altitude = 20, the equation of the base being c^' + h^'^ = 6%>. 

Ans. 8J J 1* dzdydx = 2waJbc 

4. Find the entire volume bounded by the surface (-J +{-) +(-) =!• 

. abe 
Ana. — . 
90 

6. Find the entire volume bounded by the surface x' + y' + «' = ai . . 

Ana, • 

35 

6. Find the volume cut from a sphere of radius a by a right circular cylinder 
with b as radius of base, and whose axis passes through the center of the sphere. 

8 

7. Find by triple integration the volume of the solid bounded by the planes 
z = a, y = 6, z = mx and the coordinate planes XOT and XOZ. Ana, \ mbaK 

8. The center of a sphere of radius r is on the surface of a right circular cylin- 

der the radius of whose basis is - • Find the volume of the portion of the cylinder 
intercepted by the sphere. Ana, f (r — }) r". 

9. Find the volume bounded by the hyperbolic paraboloid cz = xy, the XOr 
plane, and the planes x = ai, z = oj, y = 61, y = 62. ^ (os' — ai') (V — fti*) 

^ 4c 

10. Find the volume common to the two cylinders z' + y* = f^ and z* + 2;* = r^. 

Ana, —- — 
8 



SUCCESSIVE AND PARTIAL INTEGRATION 



419 



11. Find the volnme bounded by the plane jb = 0, the cylinder 

(X - a)2 + (y - 6)9 = r«, 
and the hyperbolic paraboloid xy = ex. Ans, 

12. Find the volume of the solid bounded by the surfaces 

« = 0, x2 + y« = 4az, z^ + y^ = 2cx. Ans. 



c 
3tc« 



8a 



13. Find the volume included by the surfaces y^-^- z^=z4az and x — z = a. 

Ana, 8 xa*. 

14. Find the volume of the solid in the first octant bounded hy xy = az and 

*+y+'=<^ 4«*. (g-iog4)«.. 

16. Find the volume included between the plane jb = 0, the cylinder y* = 2cx—x^, 
and this paraboloid az^ + 6^ = 2 2. 



16. Find the entire volumes of the solids bounded by the following surfaces : 



(c) (x2 + ya + ««)« = 27 a^zyz. 

(d) (x* + y9 + 2* + ca - o2)« = 4c»(x» + y»). 

(e) (x» + y« + «»)» = cxyz. 



Ana, 

Ana. 

Ana, 
Ana. 
Ana. 

Ana. 



36 

Swabc 

_________ . 

6 
9as 

■ • 

2 
2y«ca9. 
c« 

3eo' 

T«a6c 



4V2 



239. Miscellaneous applications of the Integral Calculus. In § 227 
it was shown how to calculate the volume of a solid of revolution 
by means of a single integration. 
Evidently we may consider a solid 
of revolution as generated by a 
moving circle of varying radius 
whose center lies on the axis of 
revolution and whose plane is per- 
pendicular to it. Thus in the 
figure the circle ACBDy whose 
plane is perpendicular to OXy may 
be supposed to generate the solid of revolution — EGFHy 
while its center moves from to JV, the radius MC (= y) varjring 




420 



INTEGRAL CALCULUS 



continuously with 0M{= x) in a manner determined by the equa- 
tion of the plane curve that is being revolved. 

We will now show how this idea may be extended to the calcu- 
lation of volumes that are not solids of revolution when it is pos- 
sible to express the area of parallel plane sections of the solid as 
a function of their distances from a fixed point. 

Suppose we choose sections of 
a solid perpendicular to OX and 
take the origin as our fixed point 
Assuming FDE as such a section, 
we have from (2>), p. 404, the area 

A= j I dzdy^ 
S 

X being regarded as constant and the limits of integration being 
extended over the area S^ {DFE), If the area of DEF is expressible 
as a function of its distance from the origin (= z), we then have 




(^) 



^ ^dzdy^fix). 
S 



But from § 238, p. 416, the volume of the entire solid is 

V=CCCdzdydx = ff f f^^^yl^^^- 



E 



S 



Hence, substituting from {A)j we have 
{B) r=Jf(x)dx, 

where f(x) is the area of a section of the solid perpendicular to 
OX expressed in terms of its distance (= x) from the origin, the 
aj-liraits being chosen so as to extend over the entire region E 
occupied by the solid. 

Evidently the solid — ABC may be considered as being gener- 
ated by the continuously varying plane section DEF as ON (= x) 
varies from zero to OM, The following examples will further 
illustrate this principle. 



SUCCESSIVE AND PARTIAL INTEGRATION 



421 




Ex. 1. Calculate the volume of the ellipsoid 

Xa y2 jfl 

—+—+-=1 
a!^ l^ c^ 

by means of a single integration. 

Solution, Consider a section of the ellipsoid perpendicular to OX, as ABCD 
with semiaxes 6' and c\ The equation of the ellipse UEJG in the XOT plane is 

Solving this for y (= h') in terms 
of X (= OM) gives 

a 

Similarly from the equation of 
the ellipse EFQI in the XOZ plane 
we get 

a 
Hence the area of the ellipse (section) ABCD is 

T6r=^(a3-x«) =/(«). 

Substituting in (B), p. 420, 

y=l!2: C (a2 - z'^)dx = ^iroftc. Ans. 
a^ J-a 3 

We may then think of the ellipsoid as being generated by a 
variable ellipse ABCD moving from G to E^ its center always on 

OX and its plane perpendicular to OX, 

Ex. 2. Find the volume of a right conoid with circular base, the radius of base 
being r and altitude a. 

Solution. Placing the conoid as shown in the figure, consider a section PQR 
perpendicular to OX, This section is an isosceles triangle; and since 

RM=y/2rz-x^ 

(found by solving x^ -h y^=^2 rz^ the equation of the 
circle OR A Q, for y) and 

JfcfP = a, 

the area qt the section is 

aV2fx-x2=/(x). 

Substituting in (B), p. 420, 

F= a f V2rx-x2dx = 
Jo 




Tf^a 



Ans, 



This is one half the volume of the cylinder of the same base and altitude. 



422 INTEGRAL CALCULUS 

Ex. 3. A rectangle moves from a fixed point, one side varying as the distance 
from this point, and the other aa the square of this distance. What is the volume 
generated while the rectangle moves a distance of 2 ft.? Ans, 4^ cu. ft. 

Ex. 4. On the double ordinates of the ellipse — h t:: = 1, isosceles triangles of 

vertical angle 90° are described in planes perpendicular to that of the ellipse. Find 

the volume of the solid generated by supposing such a variable triangle moving 

from one extremity to the other of the major axis of the ellipse. . 4 ab^ 

AfU. • 

3 

Ex. 6. Given a right circular cylinder of altitude a and radius of base r. 
Through a diameter of the upper base pass two planes which touch the lower 
base on opposite sides. Find the volume of the cylinder included between the 
two planes. Ana. (w — |) ca^. 

Ex. 6. Two cylinders of equal altitude a have a circle of radius r for their com- 
mon upper base. Their lower bases are tangent to each other. Find the volume 
common to the two cylinders. . 4 r^ 

3 

Ex. 7. Determine the amount of attraction exerted by a thin, straight, homo- 
geneous rod of uniform thickness, of length 2, and of mass if, upon a material point 
P of mass m situated at a distance of a from one end of the rod in its line of 

direction. 

— « — y^l Solution. Suppose the rod to be divided 

__, J into equal infinitesimal portions (elements) 

of length dx. 

—- = mass of a unit length of rod ; 

* 

hence —dx = mass of any element. 

Newton^s Law for measuring the attraction between any two masses Is 

• < i.A ^' product of masses 
force of attraction = — ; 

(distance between them)' 
therefore the force of attraction between the particle at P and an element of the rod is 

— mdx 



m 



(x 4- a)2 

which is then an element of the force of attraction required. The total attraction 
between the particle at P and the rod being the limit of the sum of all such ele- 
ments between x = and x = 2, we have 



force of attraction = 




Mm /•? dx Mm 

= I = 4- - . Ans. 

I Jo (x + a)-» a{a-^l) 



SUCCESSIVE AND PARTIAL INTEGRATION 



428 



Ex. 8. A vessel in the form of a right circular cone is filled with water. If A is 
its height and r the radius of hase, what time will it require to empty itself through 
an orifice of area a at the vertex ? 

Solution. Neglecting all hurtful resistances, it is known that the velocity of 
discharge through an orifice is that acquired hy a body falling freely from a height 
equal to the depth of the water. If then x denote depth 
of water, . 

Denote by dQ the volume of water discharged in time 
dtf and by dx the corresponding fall of surface. The vol- 
ume of water discharged through the orifice in a unit of 

time is y 

a V 2 fifx, 

being measure d as a right cylinder of area of base a and 
altitude v (= V2gz). Therefore in time dt, 

(A) dQ = aV2gzdL 

Denoting by 8 the area of surface of v^ter when depth is x, we have, from 




Geometry, 



8 x^ „ ,wr^^. 



But the volume of water discharged in time dt may also be considered as the 
volume of cylinder AB of area of base S and altitude dx ; hence 

irr^E^dx 
(E) dQ = 8dx = —j^. 

Equating (A) and (B) and solving for dt, 

Tt*c«dx 



dt = 



Therefore 



ah^V2gx 



'i 



2Tr«VJi 



ah*V2ffX baV2g 



Ana, 



CHAPTER XXXII 



ORDINART DIFFERENTIAL EQUATIONS* 

240. Differential equations. Order and degree. A differential 
equation is an equation involving derivatives or differentials. 
Differential equations have been frequently employed in this book, 
the foUowmg being examples. 



(3) tmf^ = p. 

(4) g=12(2:r-l). 
(6) dy = -^dx. 



a'y 



(6) dp 



a"* sin 2 d 



dd. 



(7) i«y = (20«'-12a;)da?. 
^ ^ '^di? ^ dxdy dx 



(10) 



a*?* 



a^a^az 



= (1 + 3 xyz + j^i^^u. 



Ex. 1, p. 155 



(^), p. 98 



Ex.1 



Ex.2 



Ex. 3 



Ex.1 



Ex.7 



Ex. 8 



Ex.7 



p. 113 



p. 146 



p. 145 
p. 146 
p. 197 



p. 207 



p. 207 



* A few types only of differential equatlonB are treated In this chapter, namely snch as the 
student is likely to encounter in elementary work in Mechanics and Physics. 

424 



ORDINARY DIFFERENTIAL EQUATIONS 426 

In fact, all of Chapter XIII in the Differential Calculus and 
all of Chapter XXV in the Integral Calculus treats of differential 
equations. 

An ordinary differential equation involves only one independent 
variable. The first seven of the above examples are ordinary 
differential equations. 

K partial differential equation involves more than one independent 
variable, as (8), (9), (10). 

In this chapter we shall deal with ordinary differential equations 
only. 

The order of a differential equation is that of the highest deriva- 
tive (or differential) in it. Thus (3), (6), (6), (8) are of the fir^t 
order; (1), (4), (7) are of the second order; and (2), (10) are of the 
third order. 

The degree of a differential equation which is algebraic in the 
derivatives (or differentials) is the power of the highest derivative 
(or differential) in it when the equation is free from radicals and 
fractions. Thus all the above are examples of differential equa- 
tions of the first degree except (2), which is of the second degree, 

241. Solutions of differential equations. Constants of integration. 
A solution or integral of a differential equation is a relation between 
the variables involved by which the equation is identically satis- 
fied. Thus 

{A) y=^i sin ^ 

is a solution of the differential equation 

(5) , g+y-o. 

For, differentiating (A)^ 

(C) _1 =_ c^ sin a:. 

Now, if we substitute (A) and (C) in (5), we get 

— (?j sin a; -f e?i sin a; = 0, 

showing that (A) satisfies (B) identically. Here c^ is an arbitrary 
constant. In the same manner 

(2>) y = ^a cos X 



426 INTEGRAL CALCULUS 

may be shown to be a solution of (B) for any value of c^. The 
relation 

(jE) y = <?i 81^ 2: 4- Cj cos X 

is a still more general solution of (B). In fact, by giving particular 
values to c^^ and c^ it is seen that the solution (U) includes the 
solutions (A) and (2>). 

The arbitrary constants c^ and c^ appearing in these solutions 
are called constants of integration. A solution, such as {E)^ which 
contains a number of arbitrary essential constants equal to the 
order of the equation (in this case two) is called the general 
solution or the complete integral,* Solutions obtained therefrom 
by giving particular values to the constants are called particular 
solutions or particular integrals. 

The solution of a differential equation is considered as having 
been effected when it has been reduced to an expression involv- 
ing integrals, whether the actual integrations can be effected 
or not. 

242. Verification of the solutions of differential equations. Before 
taking up the problem of solving differential equations it is best 
to further familiarize the student with what is meant by the solu- 
tion of a differential equation by verifying a number of given 
solutions. 

Ex. 1. Show that 

(1) y = c\X COB log X + C9X sin log x + x logz 
is a solution of the dififerential equation 

(2) a:2g_xg + 2y = xlogx. 

Solution. Differentiating (1), we get 

(3) J? = (c, - Ci) sin log X 4- (Cs + Ci) cos log x + log x + 1. 
ox 

... (Py . .sinlogx . . .coslogx . 1 

Substituting (1), (8), (4) in (2), we find that the equation is identically satisfied. 



* It is Bhown in works on Dlif erentlal Equations that the general soltttion has n arbitrary 
constants when the differential equation is of the nth order. 



ORDINARY DIFFERENTIAL EQUATIONS 427 



Verify the following solntioDS of the corresponding differential equations. 
Differential equations Solutions 

\dx/ dx dx 



dz/ dx dx 
\dx/ dx 

4. |[(^ + »)=«(* + y)- (2|f-«»-e)[log(x + y-l) + x-c] = 0. 

dv X '^ V 

7. (z + y)« -^ = a». y — aarc tan ^ = c, 

8. «— - y + z Vx« - y2 = 0. arc8in- = c-x, 

(2x X 

^ dy 8in2x . 

9. -^ + ycosx = . y = slnx — 1 + c«-"»*. 



dx 2 

d^ dy 

^ __ jp 

dx> dx 



10. (l-X«)-^-X— -aV = 0. y = CiC«"«»*n* + Cj«-«*«««*»*. 



243. Differential equations of the first order and of the first degree. 
Such an equation may be brought into the form Mdx -f Ndy = 0, 
in which M and N are functions of x and y. Differential equations 
coming under this head may be divided into the following t3rpe8. 

Type I. Variables separable. When the terms of a differential 
equation can be so arranged that it takes on the form 

(A) f(x)dx+F{y)dy^Q, 

where f{x) is a function of x alone and F{tf) is a function of y 
alone, the process is called separation of the variables^ and the solu- 
tion is obtained by direct integration. Thus, integrating {A)^ we 
get the general solution 

{B) ff(x) dx -^fFiy) dy = e, 

where c is an arbitrary constant. 



428 INTEGRAL CALCULUS 

Equations which are not given in the simple fonn {A) may often 
be brought into that form by means of the following rule for sepa- 
rating the variables. 

First step. Clear of fractions^ and if the equation involves derivor 
tiveSy multiply through by the differential of the independent variable. 

Second step. Collect all the terms containing the same differential 
into a single term. If then the equation takes on the form 

XYdx -f X^ Y'dy = 0, 

where X, X' are functions of x alone^ and F, F' are functions of y 
alone^ it may be brought to the form {A) by dividing through by Ji ' Y. 
Third step. Integrate each part separately^ as in (B), 

Ex. 1. Solve the equation , ^ . « 



dx (1 + z^)xy 
Solution. 

First step. (1 + x«) xydy = (1 + y«) dx. 

Second step, (1 + y^)dx - x(l + z'^ydy = 0. 

To separate the variables we now divide by x(l + z^) (1 + ^)f giving 

dx ydy 



x(l4-x2) l + y* 

dx c v^y 



= 0. 



logx - i log(l 4- x2) - i log(l + y5) =C, 

log (1 + x3) (1 4- y2) = 2 log X - 2 C. 

This result may be written in more compact form if we replace — 2 C by log c, 
i.e. we simply give a new form to the arbitrary constant. Our solution then becomes 

log (1 4- x2) (1 + 2/3) = log x3 4- log c, 

log (1 4- x2) (14-1^) = log cx3, 

(1 4- x^) (1 4- y*) = cxa. Avs. 

Ex. 2. Solve the equation 

Solution, 

First step. axdy 4- 2 aydx = xydy. 

Second step. 2 aydx -\- x{a — y)dy = 0, 

To separate the variables we divide by xy^ 

2 adz , (a — y) dy . 
h = U. 

X y 



ORDINARY DIFFERENTIAL EQUATIONS 



429 



Third tiep. 



*dx 



dy 



2 a log X + a log y - y = C, 
alogx«y = C + y, 

logex2y = -+?^. 
a a 

By passing from logarithms to exponentials this result may be written in the form 

£+? 
x«y = c« «, 

£ I 
or, x*y = c« • c*». 

c 

Denoting the constant e^ by c, we get our solution in the form 

x*y = C€«. uins. 



Differential equations 

1. ydx — xdy = 0. 

2. (l + y)(ix-- (1 --x)dy = 0. 

3. (1 + x) ydx + (1 - y)xdy = 0. 

4. (x»-o«)(ly~yda; = 0. 

5. (x2-yx«)^4-3^ + xya = 0. 

dx 

6. u«(lc4-(©-a)(lu = 0. 

du_ 1-f m' 
do ~ 1 + ua" 

9. dp + /> tan ^d^ = 0. 

10. sin e cos 0d^ — cos d sin 0d0 = 0. 

11. sec>^tan0d^ + 8ec20tan^ = O. 

12. sec> e tan 0d0 + sec^ tan dd^ = 0. 

13. xydx — (a 4- x) (6 4- y) dy = 0. 

14. (1 + x'^)dy - Vl - y^dz = 0. 



15. Vl _ x2dy + Vl - ya^x = o. 

16. 3 c* tan ydx + (1 - e*) sec" ydy = 0. 



So2utum« 
y=^cz, 

(l + y)(l-x) = c. 

log xy + X - y = c. 
X — a • 



y«a = c 



X + a 



«+y . , y 

^ + log^ = c. 

xy X 



« — a = C€". 



tt = 



« 4- c 



1 — c« 
2 1* — arc tan a = c 
p = c cos 6. 
cos = c cos $. 
tan 9 tan = c. 
8in« e + sin2 = c. 
x-y = c-\- k)g(a + x)«y*. 
arc sin y — arc tan x = c. 
y Vl-x« 4- X Vl - y2 = c, 
tan y = c (1 — c*)8 



480 INTEGRAL CALCULUS 

Type n. Homogeneous equations. The differential equation 

is said to be homogeneous when M and N are homogeneous func- 
tions of X and y of the same degree.* Such differential equations 
may be solved by making the substitution 

y=vx. 

This will give a differential equation in v and x in which the 
variables are separable, and hence we may follow the rule on p. 428. 

Ex. 1. Solve the equation 

dz dx 

SoliUUm. y^dx + (x^ - zj/) dy = 0. 

Since this is a homogeneous differential equation we transform it by means of 
the suUtituUon ^^^ Hence dy = «fa + x<fo, 

and our equation becomes 

vWdx + (x^ - vx^ {vdz + xdv) = 0, 
x^wix + x8(l - V) dv = 0. 

To separate the variables divide by ox*. 
This gives ^ 

X V 

dx . rdv 



+ (lii^» = o, 



logx + logt) - «= C, 

10geVX= C + t, 

vx = cc+*' = e<^-c», 
vx = cer. 



y 

But = - • Hence the solution is 

y = cc. An^ 



EXAMPLES 

Differenlial equatUma Solutions 

1. (x + y)dx + xdy = 0. as« + 2xy = c. 

2. (x + y)(ix-f (y-x)dy = 0. log (x« + y2)4 _ arc tan - = c. 

3. xdy - ydx = Vxa + y^dx, . 1 + 2cy - c*c« = 0. 

4. (8y + 10x)dx + (6y + 7x)dy = 0. (« + y)«(2 x + y)« = c. 

* A function of x and y Ib iaid to be homogeneotu In the yariables if the result of replacing x 
and y by Ax and Ay (A being arbitrary) reduces to the original function multiplied by some power 
of A. This power of A is called the degree of the original function. 



ORDINARY DIFFERENTIAL EQUATIONS 431 

DifferejUial eqiuxtiona Sohttians 

» 

6. (t-s)dt-^tdaz=zO. te* = c. 

m, y dy y lin- 

7. X cos - • -^ = y cos - — «. xe * = c. 

y y y 

8. X cos - (ycfcc + xdy) = y sin - (xdy — ydx). xy cos - = c. 

XX X 

Type m. Linear equations. A differential equation is said to 
be linear if the equation is of the first degree in the dependent 
variable (usually y) and its derivatives (or differentials). The 
linear differential equation of the first order is of the form 

(A) %^^y=^^ 

where P, Q are functions of x alone, or constants. 
To integrate (A), let 

(JB) y = uz, 

where 2; is a new variable and u is a function of 2; to be determined. 
Differentiating (JB), 

Substituting (0) and (B) in (A)y we get 

dz du ^ 

u-^^-z-zr-^- Puz — Q^ or, 
dx ax 

Now let us determine, if possible, the function u such that the 
term in z shall drop out. This means that the coefficient of z 
must vanish, that is, , 

dx 

Then — = - Pdx, 

u 

and log^w = — j Pda;, giving 



432 INTEGRAL CALCULUS 

Equation (D) then becomes 

dz 
ax 

To find z from the last equation, substitute in it the value of 
u from {E) and integrate. This gives 

c^dz = Qef^^dx, 
(F) c^z =^Qef^^dx -f C. 

The solution of (A) is tlien found by substituting the values of 
M and z from (E) and {F) in (B). This gives 

The proof of the correctness of ( Gr) is immediately established 
by substitution in (A), In solving examples coming under this 
head the student is advised to find the solution by following the 
method illustrated above, rather than by using (6)^) as a formula. 

Ex. 1. Solve the equation 

^ ' dx x + 1 ^ ' 

Solution, This is evidently in the linear form (A)^ where 

P = — and Q = (X + 1)*. 

x + 1 

dy dz du 
Let y = uz ; then ;,^ = '^ t" + ^ ;p • Substituting in the given equation (1), 

__^ ^^- CwJ uX uX 

we get 

dz , du 2uz , ,.« 

u— + 2— - :; = (X + 1)«, or, 

dz dx l + x ^ 



/ox ^^ . /^W 2U \ , . ,vi 



dz . /du 2u 

V 

Now to determine u we place the coefBcient of z equal to zero. This gives 

du 2u 



dx 1 H- X 

du 2dx 



u 1 4- X 
logeU = 21og(I + x), 
(8) u = ci^Kd + '>" = (1 + x)8.» 

• Since log, II = log, c»^ Ci + «)* = log (1 + -f)» • log, e = log (1 + a:)*, i t follows that u =» (1 + x^. 



ORDINARY DIFFERENTIAL EQUATIONS 433 

Equation (2) now becomes, since the term in z drops out, 

u^ = (x + l)l. 
ax 

Replacing u by its value from (3), 

cte = (x + l)*dx, 
(4) z = a^+C. 

Substituting (4) and (3) in ^ = uz, we get the solution 

y = ^ii±lL + C(x+l)2. Ann. 



EXAMPLES 

DifferenUaX eqiuUiona Solutions 

1. 2-^ = (a5 + l)». 2y = (x + l)* + c(x+l)a. 

2, -^ — ^ = y = cx^-{- 



dx z X I — a a 

3. x(l - z^dy + (2x8 - l)y(ix = caMx, y = ax -\- ex Vi - x*.* 

. - xydx odx „ ^^i 

^""T + ^'^T^' y = ax + c(l +««)*. 

5. — cos £ + <sin£=l. s = sintH-c cos t. 
dt 

ds 

6. — + a cost = 1 sin 2t. ' a = sin < — 1 + c€-«*n'. 
dt 

Type IV. Equations reducible to the linear form. Some equations 
that are not linear can be reduced to the linear form by means 
of a suitable transformation. One type of such equations is 

(A) %^^y= ^'^ 

where P, Q are functions of x alone, or constants. Equation {A) 
may be reduced to the linear form in y and z by means of* the 
substitution 2 = y""^^ Such a reduction, however, is not neces- 
sary if we employ the same method for finding the solution as 
that given under Type III, p. 431. Let us illustrate this by 
means of an example. 



434 INTEGRAL CALCULUS 

Ex. 1. Solve the equation 

(1) g + ^ = alogx.y2 

ax z 

Solution. This is evidently in the form (^1), where 

P = -, Q = alog», n = 2. 
z 

Let y = U2; then -r^=u— +« — • 

dx dx dx 

Substituting in (1), we get 

dz ^ du uz , ^, 

tc-- + 2-r -i = alogX' uH*. 

dz dx z 

Now to determine u we place the coefficient of z equal to zero. This gives 

du u ^ 

du ^ dz 
u x 

logu = -logx = log-, 

z 

(3^ u = l. 

X 

Since the term in z drops out, equation (2) now becomes 

u — = alogx-u**', 
dx 

dz 

— = a logx•^(^^ 
I dx 

Replacing u by its value from (3), 

dz , z^ 

— = alogx-— , 
dx X 

dz . dx 
-=alogx.-, 

l^ a(logx)a ^ ^^ 
z 2 

(4) z=- 



a(logx)a + 2(7 

Substituting (4) and (3) in y = U2, we get the solution 

1 2 



X a(logx)2 + 2(7 
or, xy [a (log x)a + 2 C] + 2 = 0. -4n«. 



ORDINARY DIFFERENTIAL EQUATIONS 



435 



EXAMPLES 



Differential equations 

1. -T^ + xy = x'y*. 
dx 

2. ll-z^^-xy=zaxj/*. 

dx 

3. 8ya^~ay» = x + l. 

dx 

4. x^(xV + xy)=l. 
dx 

6. (y logx — l)ydx = xdy. 

6. y — coax -^ = y" C08X (1 — Bin x). 
dx 



6oiufion« 



ir' = x2 4- 1 + ce^. 



y = {cVl ~x2-a)-i. 



yt = 


a 


1 




w' 


If* 


x[(2 


l-y2)ea^c] = 


e». 


V = 


(ex + logx + 1) 


-1 

• 


y = 


tan X + sec X 




sin 2 + c 





241 Differential equations of the nth order and of the first degree. 

Under this head we will consider four types which are of impor- 
tance in elementary work. They are special cases of linear differ- 
ential equations^ which we defined on p. 431. 

Type I. The linear differential equation 



U) 



— ^ 4- P\ — 4- P'k — 4- • • • 4- l>«f/ = O. 



in which the coefficients p^^ p^^ "iPn ^^^ constants. 

The substitution of e"" for y in the first member gives 

This expression vanishes for all values of r which satisfy the 
equation 

(B) r" +;>ir- » 4-;>»^""' + •••+;>. = ; 

and therefore for each of these values of r, e"" is a solution of (A). 
Equation (B) is called the auxiliary/ equation of (A). We observe 
that the coefficients are the same in both, the exponents in (jB) 
corresponding to the ordMpjj^^e derivatives in (A), and y in {A) 
being replaced by 1. '^ ^^rts of the auxiliary equation (B) 





436 INTEGRAL CALCULUS 

are solutions of (A). Moreover, if each one of the solutions (C) 
be multiplied by an arbitrary constant, the products 

are also found to be solutions.* And the sum of the solutions (D), 
namely 

J^ay, by substitution, be shown to be a solution of (A), Solution 
(H) contains n arbitrary constants and is the general solution (if 
the roots are all different), while (0) eive particular solutions. 

Case I. When the auodliary eqvMion has imaginary roots. Since 
imaginary roots occur in pairs, let one pair of such roots be 

r^ = a-{- bij r^ = a — hi, % = V— 1 

The corresponding solution is 

y = (?/«-^*'''>* + (?je<"-*'^* 

= «*" 1^1(008 62? -f i sin 6a:) -f (•,(cos hx — i sin 6x) (f 

= e"* \ (Cy -f (?j) cos bx -f i{Cy — <jj) sin bx\^ 
or, y = e"*(-4 cos 6x + ^sin Jj:), 

where A and B are arbitrary constants. 

* Substitutlxig c^e''^' for y in {A), the left-hand member becomes 

But this vaniflhes since r^ is a root of (J9); hence Cje**!^ is a solution of (^). Similarly for the 
other roots. 

t Replacing x by ihx in Ex. 1, p. 235, gives 

«'"*= 1 + tox H 1 ••-.or, 

L2 13 Li 15 

(1) • e'^'sl + + t( bx + \; 

12 Li V L3 15 ;' 

and replacing x by - ibx gives 

[2 L3 Li L5 

(2) e *"*=»! + \lhx- 



[2 Li V L3 [6 ; 



But, replacing x by bx in (/<), (J3), pp. 235, 236, we get 

(3) cos6x=l + ••. 

12 Li 

(4) Bmbxsabx - + •••. 

13 L6 
Hence (1) and (2) become 

e**= cos bx + i sin bxj c"**** C09 bx-i Bin bx. 



ORDINARY DIFFERENTIAL EQUATIONS 437 

Case II. When the auxiliary equation has multiple roots. Con- 
sider the linear differential equation of the third order 

where p^j p^^ p^ are constants. The corresponding auxiliary equa- 
tion is 

((7) r* ^p^^p^r + jp, = 0. 

If r^ is a root of (6?), we have shown that e**** is a solution of (F), 
We will now show that if r^ is a double root of (0^), then ze""^* is 
also a solution of (F). .Replacing y in the left-hand member of 
(F) by a^»', we get 

(H) xe^^'(r,^ ^p^r^^ +p^r^ ^p,) + e'-i^(3 r^* + 2p^r, +f a)- 
But since r^ is a double root of (6?), 

and 3 r^^ -f 2p^r^ +i?a = 0. By § 82, p. 101 

Hence (ff) vanishes, and ze""^^ is a solution of (F). Correspond- 
ing to the double root r we then have the two solutions 

More generally, if r^ is a multiple root of the auxiliary equation 
(JB), p. 435, occurring s times, then we may at once write down s 
distinct solutions of the differential equation {A)y p. 435, namely 

In case a + hi and a — hi are each multiple roots of the auxiliary 
equation, occurring % times, it follows that we may write down 2 % 
distinct solutions of the differential equation, namely 

c^e"^ cos hx^ c^e"" cos 6a;, c^e"^ cos Ja:, • • ., c^'^e"^ cos hx ; 
<j/e" sin hx^ c^xe"^ sin hxy c^^e"^ sin 6a:, • • ., c/a:*"^ e" sin hx. 

Our results may now be summed up in the following rule for 
solving differential equations of the type 

where p^^ jt?j, -"i p^ are constants. 



438 INTEGRAL CALCULUS 

First step. Write daum the corresponding auxiliary equatwn 

Second step. Solve completely the auxiliary equation. 

Third step. From the roots of the auxiliary equation write dawn 
the corresponding particular solutions of the differential equation 
as follows: 

m 

Auxiliary Equation Biffbbbntial Equation 

(a) Each distinct reaH , ^. i i ^. 

\ ' y gives a particular solution e*^»*. 

root Tj J 

(6) Each distinct pair of\ , (two particular solutions 
imaginary roots a±bi J le'^cos bxj «*" sin bx. 

, ^ 4 7.. 7 . .. ( s particular solutions obtained by 

(c) Amultiple root occur- \ . . , . , . 

.. > gives < multiplying the particular solur 

nng s times I . , , ^ « 

Itions (a) or (b) by 1, x, ar, ..., af "\ 

Fourth step. Multiply each of the n* independent solutions by 
an arbitrary constant and add the results. This gives the complete 
solution. 

Ex. 1. Solve ^ - 8^ + 4y = 0. 

Solution. Follow above rule. 

First step, H-3r*H-4 = 0, auxiliary equation. 

Second step. Solving, the roots are — 1, 2, 2. 

Third step, (a) The root — 1 gives the solution c-*. 

(b) The double root 2 gives the two solutions e*', x^*. 

Fourth step. General solution is 

y = cic-* + cje*^ + Cjxe**. Arts. 

Ex.2. 8olye^-4^ + 10^-12^ + 5» = 0. 
OX* dx' ox* c2x 

Solution. Follow above rule. 

First step, r* - 4 r* + 10 r* - 12 r + 5 = 0, auxiliary equation. . 
Second step. Solving, the roots are 1, 1, 1 ± 2 1. 

* A check on the Accuracy of the work is found in the fact that the first three steps must giT8 
n independent solutions. 



ORDINARY DIFFERENTIAL EQUATIONS 439 

Third step, (b) The pair of imaginary roots l±2i gives the two solutions 

e'co8 2x, e'8in2x(a = l, 6 = 2). 
(c) The doable root 1 gives the two solutions e*, ztF. 

Fourth step. General solution is 

y = cie* + cisc€* + cse* cos 2* + CfC'sin 2z, 
or, y = (Ci + CsX + Cs cos 2 z + C4 sin 2 x) e^. Ana, 



EXAMPLES 
Differential equationa General solutions 

dhi 

1. — ^ = 9y. y = Cie** + CjC-«*. 

dhi 

2. -r^ + y = 0. y = ci sin jc 4- ct cos z. 

3. ^ + 12y=:7^. y = cic8* + c«c** 
ox^ ox 

^ ^_4^ + 4y = 0. y = (ci + cax) ^*. 

dx^ (ix 

6. ^ ~ 4-?^ = 0. y = ci + Cje*' + c«e-«*. 

cix' ox 

6, -r? + 2^-8y = 0. y = cic*^ + Cs6-*^ + C8sin2x + C4Cos2x. 

ox* dx* 

^ d^s d^s ^da ^ ... ,, , 

dt^ dt* dt 

8. ^ - 12^ + 27p = 0. p = cic8» + C9fir*9 ^ cje^^^ C4<r»^. 

JO ^ 

9. 6— + 18u = 0. tt = (Cisin2i>4-CaC082t)c8«'. 

dv^ dv 

d^y d?v 

10. —^ + 2n*-4 + n*y = 0. y = (ci + C2x)co8 rvL-\-(c^-\- C4X) sin nx. 
dx* dx^ 

„ d»« . . -r/ . t^ . <V3 

11. — - = «. 



d«» 



a = CiC* + c s ( C3 sm h ca cos 1 . 



12. ^-7— + 6fi = 0. « = Ci^« + Cae' + C8c8«. 

d<' dt 



440 INTEGRAL CALCULUS 

Type n. The linear difierential equation 

where X is a function of x aXone^ or constant^ andp^^ J9,, •••,!?. are 
constants. 

When X=z 0, (J) reduces to (A), Type I, p. 436, 

The complete solution of (J) is called the complementary fumy 
tion of (J). 

Let w be the complete solution of (J"), i.e. the complementary 
function of (J), and v any particular solution of (J). Then 

Adding, we get 

showing that w + v is a solution* of (J). 

The complete solution of (I) being w -f v, we first find the com- 
plementary function u by placing its left-hand member equal to 
zero and solving the resulting equation by the rule on p. 438. 

To find the particular solution t; is a problem of considerable 
difficulty except in special cases. For the pi-oblems given in this 
book we may determine the particular solution v by the following 
method. 

Differentiate successively the given equation and obtain, either 
directly or by elimination, a new differential equation of a higher 
order of Type I. Solving this by the rule on p. 438, we get its 
complete solution containing the complementary function u already 
found,! and additional terms. Determining the constants of the 
additional terms so as to satisfy the given differential equation, 
we get the particular solution v. 

* In works on differential equations it is shown that u + r Is the complete solution, 
t From the method of derivation it is obvious that every solution of the original equation 
must also be a solution of the derived equation. 



ORDINARY DIFFERENTIAL EQUATIONS 441 

The method will now be illustrated by means of examples. 

Note. The solution of the auxiliary equation of the new derived differential 
equation is facilitated by observing that the left-hand member of that equation 
is exactly divisible by the left-hand member of the auxiliary equation used in find- 
ing the complementary function. 

Ex. 1. Solve 

Solution. The complementary function u of (K) is the complete solution of the 
equation 

Applying the rule on p. 488, we get as the complete solution of (L) 

Differentiating {K) gives 

^ ' dx^ dx^ dz 

Multiplying (K) by 2 and adding the result to {N), we get 

a differential equation of Type I. Solving by the rule on p. 438, we get the com- 
plete solution of (0) to be 

y = ci€f + C2«-»* + csxe-'*, 
or, from (M)^ 

y =zu + c^ier^*. 

We now determine cs so that c^ze-^* shall be a particular solution v of (K), 
Replacing y in (K) by c»xe-'*, we get 

C86-**(-4 + l) = ae-«*. 

.*. — 8 Cg = a, or, Cs = — J a. 

Hence a particular solution of (K) is 

t = — \axe-^*, 
and the complete solution is 

y =zu + v = cic* + cje-'* — J axe-**. 

Ex. 2. Solve 

(P) ^ + nay = cosax. 

Solution. Solving 

(Q) S + "'^ = "' 

we get the complementary function 

(R) tc = Ci sin TUB + Ca cos nx. 



442 INTEGRAL CALCULUS 

DifCerentiating (P) twice, we get 

(5) ^+n«^ = -o»co8ax. 

Multiplying (P) by a* and adding the result to (8) gives 

The complete solution of {T) is 

y = Ci8innx + CscoBnx + Cssinox + Cicosox, 
or, y = u + Cs sin ax + C4 cos ox. 

Let us now determine cs and C4 so that cs sin ox + C4 cos ox shall be a solution of 
(P). Replacing y in (P) by Cs sin ox + C4 cos ox, we get 

(71*04 — a*C4) cos ax + (n*C8 — a^Cj) sin ox = cos ax. 

Equating the coefficients of like terms in this identity, we get 

n*C4 — a'c4 = 1 and rA:i — a^s = 0, 

or, C4 = and Cs = 0. 

Hence a particular solution of (P) is 

cos ax 



v = 



-aa 

and the complete solution is 

cos ox 
y = u + « = Ci sin nx + Cj coenx H 



EXAMPLES 

Differential eqiuxtions Complete solutiona 

I. _4-7--2. + 12y=x. y = cic8* + cac*' + -— ^. 

dx' dx 144 

d*v dhi dh/ dy 
2- 33-2-^ + 2— ^-2--^ + y = a. y = CiBinx-f C8C0SX + {c» + C4X)c*+a. 
ox* dx' ox^ ax 

d^s t 4- 1 

dt" a* 

d^ X* 

6. -4-a*y = x>. y = CiC« + c«^*» + C8sinax4-C4COBax 

dx* a* 

d^s 2 sin ^''^ 

6. — i + a*a = cosax. » = ci sin ax + cj cos ax H 

dx^ 2a 

7. ^-^-2a5^ + a«a = c«. a = (Ci + Ci<) c«' + 



d«a d« ' ' (a - l)a 



ORDINARY DIFFERENTIAL EQUATIONS .448 

Differential equations Complete aoltUicna 






10. ^' - 9^ + 20. = W.. . = ce*. + ce" + l^±^i±Ie... 

Typem. = ^^ 

where X is a function of x alone, or constant. 

To solve this type of differential equations we have the following 
rule from Chapter XXXI, p. 392 : 

Integrate n times iuccessivelt/. Each integration will introdv^e one 
arbitrary constant 

Ex. 1. Solve -4 = xe«. 

Solution, Integrating the first time, 

^=Cze'dx, 

or, T^ = xe' - e* + Ci. By (A), p, 841 

Integrating the second time, 

-^ = r«€* • dx - fc^da: +fCidz, 

-^ = zc« - 2 c* + Cix + C2. 
dx 

Integrating the third time, 

y zzCxe^dx -('2 e^dx + fCixdx +Cc^ 

= x«F - 3c« + — ^- + C^ + Cs, 
or, y = xe' — 8 e» + Cix* + cjx + c». -4n«. 

where F is a function of y alone. 



444 INTEGRAL CALCULUS 

The rule for integrating this type is as follows : 
First step. Multiply the left-hand member by the factor 

ax 
and the right-hand member by the equivalent factor 

^dy, 
and integrate. The integral of the left-hand member wiU be* 



m 



Second step. Extract the square root of both members^ separate 
the variables^ and integrate again.^ 

Ex. 1. Solve -4 + a^y = 0. 

d!ht 
Solution. Here t4 = — <'>% hence of 1^1)6 IV. 

First step. Multiplying the left-hand member by 2 — dx and the right-hand 
member by 2dy, we get 

Integrating, (^)'= " ^^ "^ ^*' 

Second sUp. -^ = VCi - a^y*, 

dx 

taking the positive sign of the radical. Separating the variables, we get 

—^=. = dx. 

VCi - aV 

Integrating, - arc sin --L: = x + Cj, 

or, arc sin —^ = ax + aC^. 



ay 
This is the same as --^ = sin (ox + aC^) 






= sin ox cos aCi + cos ax sin aCi, 81, p. 2 

Vc^ ^ . . VcJ . ^ 

or, y = cos aCt • sm ox H sm aC^ • cob ax 

a a 

= Ci sin ox + cs cos ax. Ans. 



• Since d(^V=2^^rfx. 



<dx/ dx dx* 
t Each integration introduces an arbitrary constant. 



ORDINARY DIFFERENTIAL EQUATIONS 445 

EXAMPLES 
Differential equaiuma 8olution8 

1. ^ = x»-2co8X. y = ^ + 28ln« + cix« + cax + c». 

4. — =/8intU. 9 := - ^^B\Ti rut -^ Cit -{■ c^. 

(te« lyn + n 

^ = a^y. ax = log(y + Vya + ci) + c,, or 



"^^ dxa 



y = cie^ + cac-«*. 



8 ^ = -1-. 3t = 2ai(«*-2ci)(«* + Ci)* + Cj|. 

9. ^ = -i. (ci< + c»)« + a = ciy«. 

<Pz . /^r- , Vci^c^f + 1-1 . ^ 
10. _ = c«. tV2n = Cilog +cj. 

d<» Vci V* + 1 + 1 

11 ^ = _ * . Find L having given that -- = and « = a, when t = 0. 



^~- '=>^!I("»^*"T-')-^''^^^^ 



/^ 



CHAPTER XXXm 
INTE6RAPH. TABLE OF INTEGRALS 

245. Mechanical integration. We have seen that the determina- 
tion of the area bounded by a curve C whose equation is 

y =/(^) 
and the evaluation of the definite integral 

f{x) dx 

C 

are equivalent problems (§ 209, p. 357). 

Hitherto we have regarded the relation between the variables x 
and y as given by analytical formulas and have applied analytic 
methods in obtaining the integrals required. If, however, the 
relation between the variables is given, not analytically, but as 
frequently is the case in physical investigations, graphically, i.e. 
by a curve,* the analytic method is inapplicable unless the exact 
or approximate equation of the curve can be obtained. It is, how- 
ever, possible to determine the area bounded by a curve, whether 
we know its equation or not, by means of mechanical devices. 
We shall consider the construction theory and the use of one such 
device, namely the Integraph, invented by Abdank-Abakanowicz.f 
Before proceeding with the discussion of this machine it is neces- 
sary to take up the study of integral curves. 

246. Integral curves. If F(x) and f{x) are two functions so 
related that 

(A) j.^F(x)=f(x), 

then the curve 

(B) y = nx) 

* For instance the record made by a registering thermometer, a Bteam-«nglne indicator, or by 
certain testing machines. 

t See Lea Integraphet ; la courbe intigralc et ses applications t by Abdank-Abakanowicx, Paris, 
1889. 

446 



INTEGRAPH 



447 



is called an integral curve of the curve 

(C) y^f{x).* 

The name irdegrdl curve is due to the fact that from ((7) it is 
seen that the same relation between the functions may be expressed 
as follows : 



{J>) 



rf{x)dx = F{x). 



F{0) = 



Let us draw an original curve and a corresponding integral curve in such a way. 
as easily to compare their corresponding points. 



integral curve 
y = F{x) 



original curve 




To find an expression for the shaded portion ((/M^P') of the area under the 
original curve we substitute in (A), p. 371, giving 

area O'Af'P' =C^f(x)dz, 
But from (D) this becomes 

area O'M'P' =j^'y(x)dx = [F{x)]x=,i= ^(^^i) = AfP.t 

Theorem. For the same absciasa Xi, the number giving the length of the ordinate 
of the integral curve {B) ie the same as the nuinber thai gives the area between the 
original, curve, the axes, and the ordinate corresponding to this alt^scissa. 



* This curve l8 Bometimes called the origiruU curve. 

t When Xx- O'B't the positive area O'M'RP' \a represented by the maximum ordinate NR. 
To the right of R the area is below the axis of X and therefore negative ; consequently the 
ordinates of the integral curve, which represent the algehraic sum of the areas inclosed, will 
decrease in passing from R to 7*. 

The most general integral curve is of the form 

y«F(x) + C, . 

in which case the difference of the ordinates for x= and a;» X| gives the area under the original 
curve. In the integral curve drawn C= Fif^^O, i.e: the general integral curve is obtained if 
this integral curve be displaced the distance C parallel to O K. 



4:18 



INTEGRAL CALCULUS 



The Btadent shoald also observe that 

(a) For the same abscissa Xi, the number giving the slope of the integral carve 
is the same as the number giving the length of the corresponding ordinate of the 
original curve [from (C)]. Hence (C) is sometimes called the curve of slopes of (B). 
In the figure we see that at points O, JR, T, F, where the integral curve is parallel to 
OX, the corresponding points (X, R\ T, V on the original curve have zero ordi- 
nates, and corresponding to the point W the original curve is discontinuous. 

(b) Corresponding to points of inflection Q, jS, 17 on the integral curve we have 
maximum or minimum ordinates to the original curve. 

For example, since d /x'\ 7^ 

it follows that 

(IE) 
ia an integral curve of the parabola 

(F) 



" = ? 



y = 



x« 

■ ■■■ I 

8 



Since from (F) 



I —dx = 
i) 3 




T' 



and from (E) 
it is seen that 






indicates the number of linear 



units in the ordinate M^P^\ and also the number 
of units of area in the shaded area OMiPi. 

Also since from (E) -^ = — , or tan r = -^1 

dx 3 3 



iriPi = y. 



and from (F) 

Xi* 

it is seen that the same number -^ indicates the 
length of ordinate MiPi and the slope of the tangent at Pi'. ^ 

Evidently the origin is a point of inflection of the integral curve and a point 
with minimum ordinate on the original curve. 

S!47. The integraph. The theory of this instrument is exceedingly 
simple and depends on the relation between the given curve and a 
corresponding integral curve. 

The instrument is constructed as follows. A rectangular carriage C moves on 
rollers over the plane in a direction parallel to the axis of X of the curve 

y =/(«). 
Two sides of the carriage are parallel to the axis of X; the other two, of course, 
I)erpendicular to it Along one of these perpendicular sides moves a small carriage 
Ci bearing the tracing point T, and along the other a small carriage d bearing a 
frame F which can revolve about an axis perpendicular 'to the surface, and carries 
the sharp-edged disk D to the plane of which it is perpendicular. A stud Si is 
fixed in the carriage Ci so as to be at the same distance from the axis of X as ia 
the tracing point T, A second stud 8^ is set in a crossbar of the main carriage C 



INTEGBAFH 449 

ao u to be on tbe axis of X, A split mler B joins thrae two Btud« and slIdM 
upon thetD. A croBshead H elidea upon this mler and is joined to tlie frame F 
b; a parallelogTam. 

Tbe essential part 
of ttie Instrnment con- 

sista of the sharp- 

edged disk D, wUdi 
moTes under premure 
over a smooth plane 
surface (paper). Tliit 
disk wiU not slide, 
and hence as it rolls 
mmit always move 
along a path the tan- 
gent to nbicb at every 
point is tbe trace of 
the plane of tbe disk. 
If now this disk U 
caused to move, it la 

evident from tbe figure that the construction of the machine insures that the plane 
of the diak D shall be parallel to the ruler R. But if a Is the distance between tba 
ordlnatea through tbe studs 8,, Si, and r is the angle made by B (and therefore 
also plane of disk) with tbe aiis of X, we have 

{A) tanr = !!; 



and if 




v- 


= F(KO 


ia thecnrre 


traced by the point of 


contac 


of the disk, 


(B) 




tanT 


= ^.» 



Comparing {A) and (B), a ' "''' 

(C) y- = ^ Jvdi = \jf{') ^ = H^')-^ 

That is (dropping the primes), the curve 
to an integral eune of the curve 

The feictoT - evidently fixes merely the scale to which the integral curve la drawn 
and does not affect \i6form. 

A pencil or pen is attached to the carriage Ci in order to draw the curv< 
V — F(x). Displacing the disk D before tracing the original curve is equivalent U 
changing tbe constant of integration. 

ay' ilu' dx dt 
• Slnie x-x' + d. where d = width of niBObLne. and tberetorc ,7^ = ^ ' ^ " ^ ■ 

t It !■ auamed tbmt the Initrnment !■ ao conalructrd that the alMctsMU ol an; two eolreipoad 
lug polnu of the two curve! differ 0DI7 b; b coDtlant ; hence .c l« a faoetlon of x'. 



460 



INTEGKAL CALCULUS 



248. Integrals for reference. Following is a table of integrals 
for reference. In going over the subject of Integral Calculus for 
the first time, the student is advised to use this table sparingly, if 
at all. As soon as the derivation of these integrals is thoroughly 
understood, the table may be properly used for saving time and 
labor in the solution of practical problems. 



SOME BLSMSNTASY FORMS 



1. f{du ±dv±dw±"') = ftiu ± Cdv ± Cdw ± 



2. fado = afdv. 



jc-dx = -—^ + C, n ± 1. 



n + 1 



'dx 



3. Jdf(x)=ff(x)dx^f(x)-^C. 5. J— = log« + a. 



Forms containino Integral Powers of a + 6x 



•/ a + ox 



(a + 6x)» + i 



J o(n -f- 1) 

8. Cf{x, a-\-hx)dx. Try one of the substitations, z = a + &e, xz = a + te. * 

/xdx 1 
^—^ = -[a + te - a log(a + te)] +C. 

/x^dx 1 
-— — = -[i(a + te)» - 2a(a + 6x)+ a«log(o + 6x)] +C. 
a A- ox cp 



/ dx -. _1 1 
2 (a + bx) a 



1 , a + bx 



(a + bx) 

dx 
(o+te) 



+ C. 



1.6, a 4- 6x 



2. r_?5_=..-L+^iog 

Ja»(a+ew) ax o» 
3. A-^, = i riog(a + te) + ^-1 + C. 

^ (a + 6x)» 6" L o + teJ 



TABLE OF INTEGRALS 461 



15. r_*L^=_^ liog^±^+c. . 

16. r ^ =ir-_L,^_g ]+c. 

J(o + 6x)» ft^L a + te 2(a + te)aJ 

FoBMB coKTAiNiMo tt^ + x^ fl^ ~ X*, a + 6x", a + &^ 

J a^-{-x^ a a J 1 + x« 

Ja^-xa 2a *a-x J x* - a« 2a *x + a 

/dx 1 /6 
r-;: = — ^^tan-^x-v/- + C, when a>0 and &>0. 

Ja2-6^* 2a6 ^a-bx^ 

21. rx"»(a + 6x")Pdx 

x'"-»+i(o + 6x«)i» + » a(m-n + l) /• , .. v ^ 
6(np + m + l) 6(np + m + l)J ^ ' 

22. r«-(a + bx^)Pdx = ^■■^'(a + te')p any r ^ 
J np + m+1 np + m + lJ 

23. r — * 

^x^CaH- 

+ np — 1) /• cte_ 

— l)o Jx'»-"(a + 



dx 
(a + 6x")i» 

1 (m — n + np — 1) 6 /• dx 



24. r ^- 

^x"«(a + 6x 



(m — l)ax»"->(a + 6x»)i»-i (m — l)o ^ x'"-"(a + te»)i» 

(a + 6x")p 

1 m — n-^-np — l r ^c 

an(p-l)x«-i(a + &x*»)p-i an(p - 1) J x«(a + &x»)p-i 



25 r (q + &x»)Pdx __ _ (g-i- to')i»-<-* __ h(m - n -- np ~ 1) /" (a + 6x»)Pdx 
J x"» "" a(m — l)x*»-i a(m — 1) J x"«-" 

26 r (q -f 6x")Pdx _ (g + &x»)p anp /" (a + 6x")>~'dx 
J x*" (np — m + l)x"«-i np — m4-l«^ x"* 

27 r_?!!!^5__ - x»-* + * a(m ~n + l) r x^-^dx 

' J (a + 6x»)p " 6(m - np + 1) (a + 6x»)p-i ~ 6{m - np + 1) J (a + 6x»)p 

28 r ^"^fa? _ a?""*"^ m + n — np -f 1 / ■ x*Hix 

J (g + 6x")p ■" gn(p - 1) (g + 6x»)p-i ~ gn(p-l) J (g + 6x«)P-i' 

29 f ^ = t r g . .2n _ 3\ r <^ 1 

J (a« + x2)» 2 (n - 1) a" L(aa + x«)»-i ^ V (g* + xa)"-0 ' 



452 INTEGRAL CALCULUS 

30 r_^_-_i_r - + (2n-8)r ^ ^1. 

J (a^ te«)» " 2 (n - 1) a L(a + te«)— * ^ V (a + te«)— U 



, where z = «*. 



31 r_^^^_=if_ 

J (a + 6x«)« " 26(n - 1) (a + te^)*-! 26(n - 1) J (a + te^)»-i* 

33. f— ^L_ = ±iog-^+C. 
J * (a + 6x») on o + fee" 

Ja + bx^ 26 \ 6/ 



^^ /• x%B _ 5 __ a / * (fa; 
J a + te« "" 6 "" 6 J a + to* 

OP* /• <fa5 1 I ** . /^ 

dx 



38. f— ^ = -1-5 f_« 

39 f ^ = ^ + -L f ^ 

* J (a + &x3)2 2a(a + 6x2) 2a J a + 6x2' 



Forms containing Va + 6x 



40. rxV^Ttod» = - ^<^''-«^>J<^+^ +C. 
•/ 16 0* 



41. r..vsTte<te=^<«^iii^^^±l|^>^5«±^+o. • 

•/ 1066* 

42. r_^= = -?i^^i:^V^i:te+c. 

•^ V^Tte 362 

^^ r x^dx 2(8a2-4a6x + 862x2) / ;- ^ 

43. ( "^^^ _,M^ ^ W a + 6x -f C. 

•^ VaT6x 15^ 

-. /• dx 1 , Va 4- 6x — Va ^ . ^ 

44. I — ■ =--log — --^-^ --ifC, fora>0. 

•^ X V o + 6x V a V a + 6x 4- V a 

^.f /• dx 2 ^ , /o 4- 6x . ^ - ^ 

45. I — = tan-i^— I- hC, fora<a 

•^ x Va + 6x v-a ^ ~* 



TABLE OF INTEGRALS 453 



AR C ^ — — Vg + to 6_ r dx 

47. r^5±^ = 2vnto4-ar— ^=. 

•^ * -^x vo + te 

FOBKS CONTAININO VjC» + a* 

48. n^a + a«)*dx = ?Vx» + a?» + ^log(x + Vxa + o«) + C. 
•/ 2 2 

49. 0x2 + a2)«dx = 5(2x2 + ba^ Vx2 + a* + ?^log(x + Vx^ + a^) + C. 
9/0 8 

60. r(x» + a»)»(to = ^^±^' + J!^ r(«« + ««)'-' <te. 



• + 2 



61. Cx (x* + a«)2dx = ^^' "*""!? ' + C. 
J n + 2 

. rx»(x« + a2)*dx = f (2x2 ^. a2) Vx?» + a* - ^log(x + Vx* + a*) +C. 

•/ o 8 



62 

dx 



. f - = log(x + Vx2 + a2)+C. 

•^ (x2 + a2)* 



63 

•'(xa 

dx 



64. /-^^ = _^==+C. 
•^ (x2 + a2)« o2 Vx» + a* 

56. f ^'^ , ==Vx2-f a2 + C. 
•^ (x2 + o2)* 

66. f ^'^ , =^Vx2 + a2---log(x + Vx2>fa2)>fC. 
•'(x2 + a2)* 2 2 *' 

67. r_?!^ = -_J== + iog{x + ViMn5)+a. 

•^ (x2 + a2)* vx2 + a2 

•^x2(x2 + a2)* «^ 

-^ /• dx Vx2 + a2 . 1 , a + Vx^ + o* . ^ 

60. 1: = ~r- + — I log ^ — +C. 

61. ( ^ ^^ =Va2 + x»-alog ^ . +C. 

•/ X X 

•/ x^ *^ 



454 INTEGRAL CALCULUS 



Forms contaimino Va« — a* 

63. r(x« - a«)*daj = ? Vx^ - a^ - -logCx + Vx^ - a^) +C7. 

64. r(x« - a«)»dx = 5(2aJ« - ha^ Vx« - a« + — - log (x + Vx^ - o») + C. 

J n + 1 n + 1*^ 



«+« 



x(x2 - a«)«ix = ^= -4 — +C'. 

67. rx«(xa _ aa)*dx = ^(2x^ - a«) Vx^ - a^ - — log(x + Vx« - a^) +0. 
•/ 8 8 



68. f — ^— - = log (X + Vx* - a«) + C. 
•^ (xa - a*)* 



69. r-^-^= ^^=+C. 

•^ {x2 - o«)' aa vx2 - o« 
70. f— ?^?_= Vii3^+C. 



71 






. f— ^L??— = = Vi^3^ + - log (x + Vx« - a2) + C. 
•^(x^-a^)* 2 2 



72 ' ^ 



. f-^^ = --=L= + iog(x+v^^^:^+c. 

•^ (x« - a«)' Vx* - a2 

73. f_-^ = lsec-i?+C7; r_^= = 8ec-ix+a 
•^x(x2-a«)* « * '^ X Vx2 - 1 

/dx Vx'* — a^ ^ 

x»(x2-a2)* ^^ 



dx Vx2 - a* . 1 ,x 

^^ ^ 86C"" — 

(*«-o«)* 2a»x« 2o« a 



76. f— i? ;= Tr."' + r^.8ec-'ig+C. 



J X X 



77. I ^^ —^ = + Iog(x + vx« - o«) +0. 

•/ X* X 



TABLE OF INTEGRALS 466 



FOBXB CONTAIHINO Va* — 7^, 



a* . ,x 



78. f{a« - x«)*(i* = ? Va* - x« + ~ sin- 1 ? + C. 

79. fiaa - a!?)«daj = ? {5o« -< 2«a) V^T^a 4. ?f*sin-i- +C. 
•/ 8 8 ct 

J^ n + 1 n+lJ^ 



•i-J-2 



81. rz(a« ~ gg)«(fa = ~ ^^^ ^\^ -f C. 
J 11 + 2 

82. fx^a* - «^**«; = |(2x» - a^) Va^ - x^ + ^sin-i? +0. 
•/ 8 8 a 

83. f— ^ = Bln-i ? ; r_^ = sin-i x. 
•^ (a« - x«)* « -^ Vl-xa 



dx X 

(a* - x»)' «* Va2-x» 



84. f— ^^ = -,,-,^=. + 0. 
•'(aa- 



86. f ^'^ =-Vo«-x«+C^' 
•^ (a" - x*)* 

__ , x«dx X 

00. 



(aa-x2)* 2 2 a 



/XnXX X . ,X ^ 

-^^ = 8in-i- + C. 
^a« - x*)' Vaa - x« « 



^^ /• x»'Hfx x*""* /-^ ^ . (m-l)a2/- x«-« . 

88. I r = va^ - x2 + ^^ ^— I -(te. 

89. r ^ =llog ^= + C. 

•^x{o2-xa)* ^ a + Va2 - x* 

dx Va2 - x" 



/ox V a* — X* ^ 

xa(a«-x2)* «*^ 



/dx Va* — X* 1 , X ^ 
= = — -^ — rlog == + C. 
x»(a2--x2)* 2o2x2 2a» a ^. Va* - x* 

J X X 



93. 



'(aa _ x«)* , Va2 - x2 



^ -dx = 8in-i - _j. (7. 

x^ X a 



456 



INTEGRAL CALCULUS 



, Forms containino V2 ox — x^, V2ax + a^ 

94. CV2az-z*dx = 5-=^ V2 ox - x^ + ^ver8-»- +C. 
J 2 2 a 



96 



/: 



dx 



= vera- 



a ./ ■ 



dx 



= ver8-*«+C. 



V2ax-x« aV V2x-x« 

96. |x'»v2ax--x^dx = ^ — '--{■' -^^- I x*-iv2ax-x«dx. 

J m + 2 m + 2./ 



97 



98 



/ 



dx 



V2 ox - x« . m-1 



+ 



dx 



-l)aJa;m-iV2ax-x^ 



/x"'dx _ _ x**-^ V2qx — x^ (2m-l)o /■ x*-*dx 
V2ax-X> *» m J V2ax--x« 



xm V2 ox - X« (2m-l)ax"» (2m 
x»"dx 



99. 



■ / 



V2ax~x8 
x* 



dx = - 



(2 ax ~ x*)* m-3 /' V2ax--x» 

(2m-3)ax'» (2m-3)aJ x"-i 



00 



. fx V2ax-x*dx = 



3a« + ax-2x« ys ^ . «' i« 

1— _ v2ax - x« + — vera-i — 

6 2 a 



01 



/: 



dx 



X V2 ax - X> 



V2ax-x» 
ax 



+ C. 



xdx 



02. f ^^__ - = - V2ax~x« + OYere-^-+C. 
•^ V2ax-x2 ** 

/- x^dx X + 3 a ys ;; . 3 « , x . r* 

03. I ■ = v2ax-x« + -a«vera-»- + 0. 

•^ V2 ox - x« 2 2 a 



04. f-^^!? — ^dx = V2ax-x2 + a vera-i? +C. 
J X a 



05 



■ / 



V2 ax -- x» 
xa 



2V2ax-x« ^,^_,«.^ 

ox = vera * — hv7» 

X a 



06. r^^2^^^»^^_(2ax-xV 
J x« 3ax» 



07./ 



dx 



X — a 



08 



/ 



(2ax-x2)* aaV2ax-x2 
xdx X 



(2 ox - x2)* aV2ax-X'« 



+ C. 



+c. 



09 



. /f(x, V2S^37»)dx =/f(z -f a, V5r:^)dz, when. , = x - a 



TABLE OF INTEGRALS 467 



110 



/———^— = log{x + a + V2ax + x2) +C. 
V2 ox + aj2 

111. ri^{x, V2ax + x«)dx =fF(z - a, V2a_aa)d2, where « = x + a. 



Forms containino a-\-bx ±cz^ 



112. f ^ - = -=l=tan-i-4^i=L+C, when 6a<4ac. 

J a + bx-hcx^ y/4^ac - 6^ vToc-^ 

^^^' I . v . • g = , log , + C, when b^>i 

•/a + 6x + cx2 V6a~4ac 2cx + 6+ V6a-4ac 



ac. 



V^T7ac + 2cx-6 



114. r /^ =— J Ing ^^^-^^ +J^ZJ^r. 

•/a + to-cxa V62 + 4ac V62T4ac-2cx + 6 

116. ( ^ — = -^ Iog(2cx + 6 + 2 V^Va + 6x + cxa) + C. 

•^ Va + 6x + cx^ vc 

116. fVa + &x + cx«dx 

__ 2 ex -1-6 Va + 6x + cx« - ^""^^ 1 og (2 ex + 6 + 2 Vc Va + 6x + cx^) + C. 
4c 8c> 

117. I ^ = = -7^Bin-^ : + C. 
•^ V a + 6x - ex* V c V 6^ + 4 ac 

118. rVa + 6x-ex«dx = ^^Va-h6x~cxa + ^±igg8in-i ^^.^Zi + C. 
•^ 4c 8c» V6a+4ac 

119. C ^jg_ =^ ^+^+^' -Aiog(2cx+6-H2V^cV a+to-Hcx> )4-a 
•^ Va+6x-|-cx« c 2c« 



-^^ /• xdx Va 4- 6x — ex* 6 . , 2cx — 6 ^ 
120. I ^ — = — ^ + -- sin- > ^ + C. 

•^ Va + 6x - cx« c 2c« VP + 4ac 



Other ALGSBRiiic Forms 

121. fx/^^t^dx = V(a + X) (6 + x) + (a - 6)log(Va + x -J- V6 + x) +C. 
J \6 + X 

122. fx/^-^^B = V(a-x){6-Hx) + (a -h 6) sin-i \P±| -h C. 
•^ \6 + X \a + 

123. J^^i|dx = - V(a + X) (6 - x) - (a -h 6) sin-i^^^ + C. 



458 INTEGRAL CALCULUS 

24. Cj}--±J^dx = - VT^^ + 8in-ix +a 

26. f— =^_ = 28in-iJ^ + C. 
•^ V(x - a) {/3 - z) \^-o 

EXPOKBNTIAL AND TbIOOWOMSTRIC ForMS 

26. ra«dx = -^ + C. 129. fain xdx = -coax +C. 
J log a •/ 

27. rC(to = e' + C. 130. fcoaxdx = 8iux + C. 

28. (ef^^- \-C, 131. rtanxcix = logaecx= -logco8X+C. 

32. foot xdx = log ain x -f C 

33. faecxdx = f = log(8ecx 4- tanx) = logtaxi( - + - )+C. 

J J coax \4 2/ 

J* /• (2x X 

•coaecxdx = I = log(coflecx — cotx) = log tan- +C. 
^ am X 2 

36. Jaec^ xdx = tan x + C. 138. jcoaec x cot xdx = — coaec x + C 

36. rcoaec«xdx= -cotx + C. 139. r8in«xdx = - - -sln2x +C. 

37. faec x tan xdx = aec x + C 140. jcoa^ xdx = - + - ain 2 x + C 

-- /• . J ain"-! x coax ^ _ i /• 

41. I am* xdx = 1 I am*-* xdx. 

J n n ^ 



coa"-*X8inx n — 1 



M^ r J coa»-*X8mx . n — i /• „ , 

42. I coa» xdx = H I cos"-^ xdx. 

•/ n n J 

/• dx _ __ 1 cos X n — 2 /• dx 
Jsin^x n — l8in"-*x n— iJain'-^x 

r dx _ 1 a inx n — 2 /* dx 
J coa" X n — 1 coa'»- * x n — 1 J ( 



co8»-2x 
C08*"-*x8in" + *x m — \ 



A^ C «. • - J C08»"-»x8in"-^'x .m — 1/* _, i.j 
46. I C08™ X am" xdx = H I coa*"-* x sin" xdac 

/. , 8in»->xco8* + !x . n — 1 /• , . , 

coa"« X ain" xdx = 1 | coa* x ain"-* xdx. 



46 



. ,y /• dx _ 1 1 ^m + n — 2/* dx 

J 8in'"xco8"x n — 1 ain"*- ^x coa" -'x n — 1 •/ aii 



8in"»xco8"-*x 



TABLE OF INTEGRALS 



459 



48. 



Si 



dx 



6m"*2C08^X 



'C0S"*2dX 



m — 1 8in"«->a5C(M^-ix 



m + n - 2 r dx 
m — 1 •'Bin"«-'x< 



xcos"x 



C08* + ^X 



^q rcos^xox _ co8"-^'x m -• n + :s rcofiFa 

J 8in"x (n — l)8in»-ix n — 1 J 8ill»- 



m -• n + 2 /•cofiFxdx 

«x" 



50 



/COS 



C08"*X(IX 



cos*-*x 



m — 1 rcoflF-*x(ix 



8in»x 



8iii»x (m — n) sin* - 1 x m 

61. I 8inxco8"xdx = \-C, 

J n + 1 

CO r • - -. 8in»+ix . ^ 
52. I 8m"x cos X(2x = H C 

^ n + 1 
tan"xdx=: j tan^-^xdr +0. 

cot»xdx = Jcot»-»xdx + (7. 

EC r • J J 8in(m4-n)x , sin (m — n)x .^ 
66. I sin Tnx sin nxdx = ^^ ^ + — ^^ ^-^— +C. 

J 2(m + n) 2(m-n) 

66./ 



COS mx COS 



, sin (m + n) X . sin (m — n) X . ^ 

nxdx = — ^^ '— H ^ f- + C. 

2(m + n) 2(m-n) 



Ew r . J cos (m + n) X cos (m — n) x . ^ 

67. I sin TOX cos nxdx = ^^ — -^— -^! ^^ *— + C. 

./ 2 (w + n) 2 (m - n) 

68. r ^ =_^__ tan-if\P^ tan?) + C, when a>6. 

•/a + &co6X Va2 — \fi \ifa + 6 2/ 



69 



•/; 



dx 



1 



V6-atan- + V6 + a 
log — — -^ ^ :^:r^:: + C?, when a < 6. 



a + 6cosx V^r^a VftH^ tan ? - VdT^ 

2 



a tan - + & 

/dx 2 2 
= tan-i — -f-fi whena>5. 
o + 6sinx Va^-fta Va^-ft^ 



atan? + 6~V6a-a« 



/dx 1 2 
; = log hC, whena<& 
a + 6sinx V^TT^ a tan ? + 6 + VP^T^ 

2 

62. r ^ = ltan-ir^^?^) + C. 

^ a2cos2x + 6«sin«x a6 \ a / 

63. re^^sinnxdx = ^<^^^°^-^^^"^>+C; 

/^ , , c*(8inx — cosx) . -, 
e^sin xdx = — ^^ ^ +C. 
2 



460 INTEGRAL CALCULUS 

^aA T— * J e«*(nsiniwc + acosnx) ^ 
164. \ ef^coBTUcdz = — ^ --hC: 

166. fxc«»%te=^(a«-l)+C 

166^ ra?»e«dx = 5!^-? ra?»-ie«yix. 

167. fa^'^'T?^ = ^""^^ ? — ra»«'x"-icfe. 

•^ m log a m log a J 

168. C^L^z= g* logg f q""^ 

•^ a?" (m — l)x»"-i m — lJa;"*-i 

169. re^co6»xdx = ^^^"''^<^"^^^ + ^«'"^)^!^(!Lli)f^ 
•^ a« + n* a« + n* ^ 

170. J'x'^cos oxdx = ?^ (ox sin ox + m cos ax) - ^^'""^^ fx^-^cos axdz. 

Logarithmic Forms « 

171. jlogx<ix = xlogx-x+C. 



172. Jj— = log{logx) + logx + i logax + . • .. 



174. rx-logxdx='x» + iri5?^ 1 1+C. 

J Ln + 1 (n + l)sj 

176. /^'logx(te = ?:!^-l J^dx. 

176. f x«log"x<ix = ^^ log"x ^ fx^loff^-ixdx. 

177. f?"^ ~ _ g"*-^^ m + 1 r x^dx 
J log«x (n - l)log«-ix n - 1 J ] 



log"-ix 



INDEX 



[The niunbers refer to the pages.] 



Abaolute conTergence, 227 ; Talue, 9. 

Acceleration, 106. 

Approximate formulas, 240. 

Archimedes, spiral of, 283. 

Areas of plane carves, polar coordinates, 

376, 406 ; rectangular coordinates, 371, 

402. 
Areas of surfaces, 888, 411. 
Asymptotes, 262. 
Auxiliary equation, 436. 

Binomial differentials, 384. 
Binomial Theorem, 1, 111. 

Cardioid, 282. 

Catenary, 281. 

Cauchy's ratio test, 224. 

Change of variables, 162. 

Circle of curvature, 188. 

Cissoid, 280. 

Computation by series of e, 237 ; of 

logarithms, 238 ; of ir, 238. 
Concave up, 123; down, 124. 
Conchoid of Nicomedes, 281. 
Conditional convergence, 227. 
Cone, 2. 

Conjugate points, 264. 
Constant, 11 ; absolute, 11 ; arbitrary, 

11 ; numerical, 11 ; of integration, 

289, 309. 
Continuity, 22 ; of functions, 22. 
Convergency, 219. 

Codrdinates of center of curvature, 184. 
Critical values, 120. 
Cubical parabola, 280. 
Curvature, center of, 183; circle of, 

183 ; definition, 169 ; radius of, 169. 
Curve tracing, 266. 
Curves in space, 271. 



461 



Cusp, 263. 
Cycloid, 96, 281. 
Cylinder, 2. 

Decreasing function, 116. 

Definite integration, 866. 

Degree of differential equation, 426. 

Derivative, definition, 39. 

Derivative of arc, 141, 143. 

Differential coefficient, 39. 

Differential equations, 424. 

Differential of an area, 866. 

Differentials, 140. 

Differentiation, 37 ; of constant, 48 ; of 
exponentials, 60; of function of a 
function, 67 ; of implicit function, 84, 
260 ; of inverse circular functions, 70 ; 
of inverse function, 68 ; of logarithm, 
* 68, 62 ; of power, 61 ; of product, 60 ; 
of quotient, 62; of sum, 49; of trig- 
onometrical functions, 67. 

Double point, 262 ; integration, 396. 

Envelopes, 208. 
Equiangular spiral, 288. 
Evolute of a curve, 187, 218. 
Expansion of functions, 23L 
Exponential curve, 284. 

Family of curves, 208. 

Fluxions, 37, 44. 

Folium of Descartes, 282. 

Formulas for reference, 1. 

Function, algebraic, 17; continuity of, 
22 ; definition, 12 ; derivative of, 39 ; 
explicit, 16; explicit algebraic, 17; 
exponential, 17; graph of, 24; im- 
plicit, 83, 202 ; increasing, decreasing, 
116 ; inverse, 16, 18 ; logarithmic, 17 ; 



462 



INDEX 



Function, — continued. 
many>Yalued, 14; of a function, 67 
rational, 16; rational integral, 16 
transcendental, 17 ; trigonometric, 17 
uniform, one-valued, 14. 

Greek alphabet, 4. 

Helix, 272. 

Homogeneous differential equation, 430. 
Huygens* approximation, 242. 
Hyperbolic spiral, 283. 
Hypocycloid, 282. 

Increasing functions, 116. 

Increments,* 37. 

Indeterminate forms, 172. 

Infinitesimal, 21, 141. 

Infinity, 21. 

Inflection, 136. 

Integer, 7. 

Integral curves, 446; definite, 356; 

definition, 288; indefinite, 289. 
Integraph, 448. 
Integration, by partial fractions, 316; 

by parts, 341 ; by rationalization, 320 ; 

by transformation, 337; definition, 

287. 
Interval of a variable, 11. 
Involute, 180. 

Jordan, 248. 

Laplace, 37. 

Leibnitz, 44; formula. 111. 

Lemniscate, 281. 

Length of curves, 378. 

Lima9on, 283. 

Limit, change of, 366; infinite, 360; 

interchange of, 368; of a variable, 17; 

of integration, 167 ; theory of, 10, 26. 
Linear differential equation, 431. 
Litmus, 283. 

Logarithmic curve, 284 ; spiral, 283. 
Logarithms, Briggs*, 240 ; common, 240 ; 

Naperian, 240 ; natural, 34. 

Maclaurin^s theorem and series, 234. 
Maxima and minima, 116, 136, 160, 246. 



Mean value, extended theorem of, 168; 

generalized theorem of, 172 ; theorem 

of, 167. 
Moment of inertia, 408. 
Multiple roots, 100. 

Newton, 37, 44. 

Node, 262. 

Normal, 00, 276; plane, 271, 277. 

Numbers, complex, 0; imaginary, 8; 

irrational, 8 ; real, 8. 
Numerical value, 0. 

Order of differential equations, 426. 
Ordinary point, 269, 273. 
Osculation, 263. 
Osgood, 220. 

Parabola, 281 ; cubic, 280 ; semicubical, 
280 ; spiral, 284. 

Parameter, 11, 208. 

Parametric equations, 02. 

Partial derivatives, 103; integration, 304. 

Pierpont, 250. 

Points, conjugate, 264; end, 266; iso- 
lated, 264 ; of inflection, 136 ; salient, 
266; singular, 250; turning, 118. 

Probability curve, 284. 

Quadratic equation, 1. 

Radian, 17. 

Radius of curvature, 162. 

Rates, 148. 

Rational fractions, 815. 

Real number, 8. 

Reciprocal spiral, 283. 

Reduction formulas, 344. 

Rollers Theorem, 166. 

Semicubical parabo^ 280. 

Series, alternating, 226 ; arithmetical, 1 ; 
convergent, 210; definition, 217; diver- 
gent, 220; geometrical, 1; infinite, 
218 ; nonconvergent, 220 ; oscillating, 
220 ; power, 228. 

Sine curve, 284. 

Singular points, 269. 



INDEX 



463 



Solation of difierential equations, 426. 

Sphere, 2. ^ 

Stirling, 235. 

Strophoid, 283. 

Subnormal, 09. 

Subtangent, 90. 

SuccesBiye differentiation, 109, 146, 206 ; 

integration, 892. 
Surface, area of, 888. 

Tangent curve, 284. 

Tangent, to plane curves, 89 ; to space 

curves, 271. 
Tangent line to surface, 271. 
Tangent plane, 278. 



Taylor*s Series, 232 ; theorem, 232. 
Test, comparison, 222. 
Total differentiation, 198. 
Trajectory, 98, 127. 
Transcendental function, 17. 
Trigonometric differentials, 303. 
Triple integration, 416. 

Vall^-Poussin, 31. 

Variable, definition, 11 ; dependent, 12; 

independent, 12. 
Velocity, 102. 
Volumes of solids, 384, 416. 

Witch of Agnesi, 280. 



To avoid fine, this book should be returned on 
or before the date last stamped below 



LIBRARY, 9C 
FEB 1 4 1939 




IHOOL OF 



JUhsJms 

MARl M952 
iAY C- 1959 



f 



EDUCATION